Trigonometry

Defining Trigonometric Ratios in Right Triangles

Trigonometry is the study of relationships between angles and side lengths in triangles. In 10th Grade, one of the most important ideas in trigonometry is learning how to describe the sides of a right triangle using ratios.

In this lesson, you will learn how to define the three basic trigonometric ratios: sine, cosine, and tangent. These ratios compare side lengths in a right triangle based on a chosen acute angle.

1. Start with a right triangle

A right triangle is a triangle with one angle equal to \(90^\circ\). The side opposite the right angle is always the longest side, called the hypotenuse.

The other two sides are called the legs. Their names depend on which acute angle you are focusing on.

Suppose you choose one acute angle, called \(\theta\). Then:

• The opposite side is the side directly across from \(\theta\).
• The adjacent side is the side next to \(\theta\) that is not the hypotenuse.
• The hypotenuse stays the same no matter which acute angle you choose.

This means that the names opposite and adjacent can change if you switch to the other acute angle.

2. The three trigonometric ratios

For an acute angle \(\theta\) in a right triangle, the three basic trigonometric ratios are defined as follows:

These are often read as:

• sine = opposite over hypotenuse
• cosine = adjacent over hypotenuse
• tangent = opposite over adjacent

A common memory helper is:

SOH-CAH-TOA

• SOH: \(\sin = \frac{\text{Opposite}}{\text{Hypotenuse}}\)
• CAH: \(\cos = \frac{\text{Adjacent}}{\text{Hypotenuse}}\)
• TOA: \(\tan = \frac{\text{Opposite}}{\text{Adjacent}}\)

3. Why these ratios matter

These ratios help us connect angles and side lengths. If we know an angle, we can compare the sides. If we know side lengths, we can find the value of a trigonometric ratio.

For example, in any right triangle with the same angle \(\theta\), the ratio \(\frac{\text{opposite}}{\text{hypotenuse}}\) will always be the same. That is why sine, cosine, and tangent depend only on the angle, not on the size of the triangle.

4. How to identify the sides correctly

Before using sine, cosine, or tangent, always follow these steps:

1. Find the right angle.
2. Label the side opposite the right angle as the hypotenuse.
3. Choose the acute angle you are working with.
4. Find the side across from that angle: this is the opposite.
5. Find the side next to that angle that is not the hypotenuse: this is the adjacent.

This labeling step is very important. Many mistakes happen because students mix up opposite and adjacent.

5. Worked Example 1: Identify the ratio

In a right triangle, relative to angle \(A\):

• opposite side = 3
• adjacent side = 4
• hypotenuse = 5

Find \(\sin(A)\), \(\cos(A)\), and \(\tan(A)\).

Step 1: Use the definitions.

Answer:

• \(\sin(A)=\frac{3}{5}\)
• \(\cos(A)=\frac{4}{5}\)
• \(\tan(A)=\frac{3}{4}\)

6. Worked Example 2: Same triangle, different angle

Now use the same triangle with side lengths 3, 4, and 5, but this time look at the other acute angle, \(B\).

Relative to angle \(B\):

• the side of length 4 is now opposite
• the side of length 3 is now adjacent
• the hypotenuse is still 5

Find \(\sin(B)\), \(\cos(B)\), and \(\tan(B)\).

This example shows an important idea: the side names depend on the angle you choose.

7. Worked Example 3: Decide which ratio to use

In a right triangle, relative to angle \(\theta\), the opposite side is 8 and the hypotenuse is 17. Find the trigonometric ratio that compares these two sides, and then write its value.

Step 1: Identify the sides involved.

The sides given are opposite and hypotenuse.

Step 2: Choose the correct ratio.

The ratio using opposite and hypotenuse is sine.

Answer: \(\sin(\theta)=\frac{8}{17}\)

8. Worked Example 4: Find a missing side ratio from side lengths

A right triangle has side lengths 5, 12, and 13. Relative to angle \(\theta\), the side of length 12 is opposite and the side of length 5 is adjacent.

Find all three trigonometric ratios.

Step 1: Identify the hypotenuse.

The longest side is 13, so it is the hypotenuse.

Step 2: Use the definitions.

Answer:

• \(\sin(\theta)=\frac{12}{13}\)
• \(\cos(\theta)=\frac{5}{13}\)
• \(\tan(\theta)=\frac{12}{5}\)

9. Important notes and common mistakes

• Do not guess the hypotenuse. It is always opposite the right angle and is always the longest side.
• Opposite and adjacent depend on the chosen angle. If the angle changes, those labels may switch.
• Tangent does not use the hypotenuse. It compares opposite and adjacent.
• Write ratios as fractions. Keep them simplified when possible.

10. Quick check for understanding

If you are given a right triangle and an angle \(\theta\), ask yourself:

• Which side is across from \(\theta\)? That is opposite.
• Which side touches \(\theta\) but is not the hypotenuse? That is adjacent.
• Which side is across from the right angle? That is the hypotenuse.

Then match the ratio you need:

• Need opposite and hypotenuse? Use \(\sin\).
• Need adjacent and hypotenuse? Use \(\cos\).
• Need opposite and adjacent? Use \(\tan\).

11. Brief summary

In a right triangle, sine, cosine, and tangent are ratios based on a chosen acute angle. The three definitions are:

The key to success is labeling the sides correctly first. Once you know which side is opposite, adjacent, and hypotenuse, you can choose the correct trigonometric ratio with confidence.

Reciprocal Trigonometric Ratios are three trigonometric ratios that are built from the three main ratios you may already know: sine, cosine, and tangent.

If you know that:

• \(\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}\)

• \(\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}\)

• \(\tan \theta = \frac{\text{opposite}}{\text{adjacent}}\)

then the reciprocal ratios are just the flipped versions of these fractions.

In math, a reciprocal means the multiplicative inverse. That means if you multiply a number by its reciprocal, the product is 1.

For example, the reciprocal of \(\frac{2}{3}\) is \(\frac{3}{2}\), because:

The same idea works for trigonometric ratios.

The reciprocal trigonometric ratios are:

• Cosecant: \(\csc \theta\), the reciprocal of sine

• Secant: \(\sec \theta\), the reciprocal of cosine

• Cotangent: \(\cot \theta\), the reciprocal of tangent

So the three reciprocal identities are:

Because sine, cosine, and tangent are ratios of side lengths in a right triangle, we can also write the reciprocal ratios directly using triangle sides.

Suppose \(\theta\) is one acute angle in a right triangle. Then:

• Cosecant compares hypotenuse to opposite:

  \(\csc \theta = \frac{\text{hypotenuse}}{\text{opposite}}\)

• Secant compares hypotenuse to adjacent:

  \(\sec \theta = \frac{\text{hypotenuse}}{\text{adjacent}}\)

• Cotangent compares adjacent to opposite:

  \(\cot \theta = \frac{\text{adjacent}}{\text{opposite}}\)

Here is the full set of six ratios together:

A helpful pattern is that each reciprocal ratio simply flips the fraction of its matching primary ratio.

For example:

• If \(\sin \theta = \frac{3}{5}\), then \(\csc \theta = \frac{5}{3}\)

• If \(\cos \theta = \frac{4}{7}\), then \(\sec \theta = \frac{7}{4}\)

• If \(\tan \theta = \frac{2}{9}\), then \(\cot \theta = \frac{9}{2}\)

Important note: a reciprocal ratio is not found by subtracting from 1. For example, \(\csc \theta\) is not \(1 - \sin \theta\). It is \(\frac{1}{\sin \theta}\).

Now let’s work through some examples.

Worked Example 1: Find a reciprocal ratio from a known ratio

Suppose:

Find \(\csc \theta\).

Since cosecant is the reciprocal of sine:

Answer: \(\csc \theta = \frac{17}{8}\)

This is a direct flip of the sine ratio.

Worked Example 2: Use side lengths in a triangle

A right triangle has:

• opposite side = 6

• adjacent side = 8

• hypotenuse = 10

Find all three reciprocal ratios for angle \(\theta\).

Start by using the side definitions:

Answers:

• \(\csc \theta = \frac{5}{3}\)

• \(\sec \theta = \frac{5}{4}\)

• \(\cot \theta = \frac{4}{3}\)

You could also first find \(\sin \theta\), \(\cos \theta\), and \(\tan \theta\), then flip each one.

Worked Example 3: Find a missing reciprocal ratio from tangent

Suppose:

Find \(\cot \theta\).

Cotangent is the reciprocal of tangent, so:

Answer: \(\cot \theta = \frac{11}{3}\)

Worked Example 4: Find secant from a triangle with a missing side

In a right triangle, for angle \(\theta\):

• adjacent side = 9

• opposite side = 12

Find \(\sec \theta\).

To find secant, we need the hypotenuse. Use the Pythagorean Theorem:

Now use the secant definition:

Answer: \(\sec \theta = \frac{5}{3}\)

This example shows that sometimes you must find a missing side before using a trig ratio.

How to remember the reciprocal ratios

• \(\csc\) matches with \(\sin\)

• \(\sec\) matches with \(\cos\)

• \(\cot\) matches with \(\tan\)

A quick memory idea is to think of them as partner pairs:

• sine ↔ cosecant

• cosine ↔ secant

• tangent ↔ cotangent

Common mistakes to avoid

• Do not confuse reciprocal with opposite in the triangle. A reciprocal means flip the fraction.

• Do not write \(\sec \theta = \frac{1}{\sin \theta}\). Secant goes with cosine, not sine.

• Do not forget to simplify fractions when possible.

• Make sure you identify opposite and adjacent based on the given angle \(\theta\).

Quick check

If \(\cos \theta = \frac{5}{13}\), then:

If \(\tan \theta = \frac{7}{4}\), then:

If \(\sin \theta = \frac{9}{10}\), then:

Summary

Reciprocal trigonometric ratios are the flipped versions of sine, cosine, and tangent.

• \(\csc \theta = \frac{1}{\sin \theta} = \frac{\text{hypotenuse}}{\text{opposite}}\)

• \(\sec \theta = \frac{1}{\cos \theta} = \frac{\text{hypotenuse}}{\text{adjacent}}\)

• \(\cot \theta = \frac{1}{\tan \theta} = \frac{\text{adjacent}}{\text{opposite}}\)

When you know a primary trig ratio, you can find its reciprocal by flipping the fraction. When working with a triangle, use the side definitions carefully based on the location of the angle.

Trigonometric Ratios of Standard Angles

In trigonometry, we study how angles and side lengths are connected in a right triangle. The three main trigonometric ratios are sine, cosine, and tangent.

For some special angles, called standard angles, the values of these ratios can be found exactly. In 10th Grade, the most important standard angles are \(0^\circ, 30^\circ, 45^\circ, 60^\circ,\) and \(90^\circ\).

Learning these exact values helps you solve many trigonometry questions quickly and accurately.

1. Review: the three trigonometric ratios

In a right triangle, choose one acute angle \(\theta\). Then the sides are named like this:

• Hypotenuse: the longest side, opposite the right angle
• Opposite side: the side across from \(\theta\)
• Adjacent side: the side next to \(\theta\), not the hypotenuse

The three ratios are:

$$ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}, \qquad \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}, \qquad \tan \theta = \frac{\text{opposite}}{\text{adjacent}} $$

A useful memory aid is SOH-CAH-TOA:

• SOH: \(\sin = \frac{\text{Opposite}}{\text{Hypotenuse}}\)
• CAH: \(\cos = \frac{\text{Adjacent}}{\text{Hypotenuse}}\)
• TOA: \(\tan = \frac{\text{Opposite}}{\text{Adjacent}}\)

2. Finding exact values using special triangles

The exact trigonometric values for standard angles come from two special right triangles:

• a 45°-45°-90° triangle
• a 30°-60°-90° triangle

3. The 45°-45°-90° triangle

If a right triangle has angles \(45^\circ, 45^\circ, 90^\circ\), then the two shorter sides are equal.

Suppose each shorter side is \(1\). Using Pythagoras' theorem:

$$ \text{hypotenuse} = \sqrt{1^2+1^2} = \sqrt{2} $$

So the side ratio is:

$$ 1:1:\sqrt{2} $$

Now find the trigonometric ratios for \(45^\circ\):

$$ \sin 45^\circ = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} $$

$$ \cos 45^\circ = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} $$

$$ \tan 45^\circ = \frac{1}{1} = 1 $$

4. The 30°-60°-90° triangle

This triangle can be made by cutting an equilateral triangle in half.

Start with an equilateral triangle of side length \(2\). When it is cut into two equal right triangles:

• the hypotenuse is \(2\)
• the shorter side is \(1\)
• the other side is found using Pythagoras' theorem

$$ \text{other side} = \sqrt{2^2-1^2} = \sqrt{4-1} = \sqrt{3} $$

So the side ratio is:

$$ 1:\sqrt{3}:2 $$

In this triangle:

• the side opposite \(30^\circ\) is \(1\)
• the side opposite \(60^\circ\) is \(\sqrt{3}\)
• the hypotenuse is \(2\)

Now we find the exact values.

For \(30^\circ\):

$$ \sin 30^\circ = \frac{1}{2} $$

$$ \cos 30^\circ = \frac{\sqrt{3}}{2} $$

$$ \tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} $$

For \(60^\circ\):

$$ \sin 60^\circ = \frac{\sqrt{3}}{2} $$

$$ \cos 60^\circ = \frac{1}{2} $$

$$ \tan 60^\circ = \sqrt{3} $$

5. Values at \(0^\circ\) and \(90^\circ\)

These values are also standard and should be memorized.

$$ \sin 0^\circ = 0, \qquad \cos 0^\circ = 1, \qquad \tan 0^\circ = 0 $$

$$ \sin 90^\circ = 1, \qquad \cos 90^\circ = 0 $$

For tangent at \(90^\circ\):

$$ \tan 90^\circ = \frac{\sin 90^\circ}{\cos 90^\circ} = \frac{1}{0} $$

Division by zero is not possible, so \(\tan 90^\circ\) is undefined.

6. Table of standard values

Here is the full table you should know:

$$ \begin{array}{c|ccccc} \theta & 0^\circ & 30^\circ & 45^\circ & 60^\circ & 90^\circ \\ \hline \sin \theta & 0 & \frac{1}{2} & \frac{\sqrt{2}}{2} & \frac{\sqrt{3}}{2} & 1 \\ \cos \theta & 1 & \frac{\sqrt{3}}{2} & \frac{\sqrt{2}}{2} & \frac{1}{2} & 0 \\ \tan \theta & 0 & \frac{\sqrt{3}}{3} & 1 & \sqrt{3} & \text{undefined} \end{array} $$

7. A quick pattern to help memory

There is a useful pattern for sine:

$$ \sin 0^\circ = \frac{\sqrt{0}}{2}, \quad \sin 30^\circ = \frac{\sqrt{1}}{2}, \quad \sin 45^\circ = \frac{\sqrt{2}}{2}, \quad \sin 60^\circ = \frac{\sqrt{3}}{2}, \quad \sin 90^\circ = \frac{\sqrt{4}}{2} $$

That gives:

$$ 0,\ \frac{1}{2},\ \frac{\sqrt{2}}{2},\ \frac{\sqrt{3}}{2},\ 1 $$

For cosine, the same values appear in reverse order:

$$ 1,\ \frac{\sqrt{3}}{2},\ \frac{\sqrt{2}}{2},\ \frac{1}{2},\ 0 $$

This is helpful because sine increases from \(0\) to \(1\) as the angle goes from \(0^\circ\) to \(90^\circ\), while cosine decreases from \(1\) to \(0\).

8. Important relationships to notice

• \(\sin 30^\circ = \cos 60^\circ\)
• \(\sin 60^\circ = \cos 30^\circ\)
• \(\sin 45^\circ = \cos 45^\circ\)
• \(\tan \theta = \frac{\sin \theta}{\cos \theta}\)

These relationships make it easier to check your work.

9. Worked Examples

Example 1: Find \(\sin 30^\circ\), \(\cos 45^\circ\), and \(\tan 60^\circ\).

Using the standard table:

$$ \sin 30^\circ = \frac{1}{2} $$

$$ \cos 45^\circ = \frac{\sqrt{2}}{2} $$

$$ \tan 60^\circ = \sqrt{3} $$

Example 2: Use special triangles to find \(\tan 30^\circ\).

In a \(30^\circ-60^\circ-90^\circ\) triangle, the sides are in the ratio:

$$ 1:\sqrt{3}:2 $$

For the angle \(30^\circ\):

• opposite side = \(1\)
• adjacent side = \(\sqrt{3}\)

So:

$$ \tan 30^\circ = \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} $$

Example 3: Find the value of \(\dfrac{\sin 60^\circ}{\cos 60^\circ}\).

Substitute the exact values:

$$ \frac{\sin 60^\circ}{\cos 60^\circ} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} $$

Divide by multiplying by the reciprocal:

$$ \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \frac{\sqrt{3}}{2} \times \frac{2}{1} = \sqrt{3} $$

Therefore:

$$ \frac{\sin 60^\circ}{\cos 60^\circ} = \sqrt{3} $$

This matches:

$$ \tan 60^\circ = \sqrt{3} $$

Example 4: Evaluate \(2\sin 45^\circ + \cos 60^\circ\).

Write the standard values:

$$ 2\sin 45^\circ + \cos 60^\circ = 2\left(\frac{\sqrt{2}}{2}\right) + \frac{1}{2} $$

Simplify:

$$ 2\left(\frac{\sqrt{2}}{2}\right) = \sqrt{2} $$

So the final answer is:

$$ \sqrt{2} + \frac{1}{2} $$

10. Common mistakes to avoid

• Mixing up sine and cosine. Always identify opposite, adjacent, and hypotenuse carefully.
• Forgetting that \(\tan 90^\circ\) is undefined.
• Confusing \(\tan 30^\circ\) and \(\tan 60^\circ\). Remember: $$ \tan 30^\circ = \frac{\sqrt{3}}{3}, \qquad \tan 60^\circ = \sqrt{3} $$
• Not simplifying correctly, especially values like \(\frac{1}{\sqrt{3}}\), which is usually written as \(\frac{\sqrt{3}}{3}\).

11. How to study and remember the standard angles

• Memorize the table one row at a time.
• Use the sine pattern \(\frac{\sqrt{0}}{2}, \frac{\sqrt{1}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{4}}{2}\).
• Get cosine by reversing the sine row.
• Find tangent using \(\tan \theta = \frac{\sin \theta}{\cos \theta}\).
• Practice writing the full table from memory.

12. Summary

Standard angles are special angles whose trigonometric ratios can be found exactly. The values for \(30^\circ, 45^\circ,\) and \(60^\circ\) come from the special right triangles \(30^\circ-60^\circ-90^\circ\) and \(45^\circ-45^\circ-90^\circ\).

You should know the exact values of \(\sin\), \(\cos\), and \(\tan\) for \(0^\circ, 30^\circ, 45^\circ, 60^\circ,\) and \(90^\circ\). These values are essential for solving trigonometry problems quickly and correctly.

Trigonometric Ratios of Complementary Angles

In a right triangle, the two acute angles always add up to \(90^\circ\). These two angles are called complementary angles.

For example, if one angle is \(35^\circ\), then the other acute angle must be \(55^\circ\), because

This simple fact leads to an important set of trigonometric relationships. These relationships help us rewrite one trigonometric ratio in terms of another, which makes many questions much easier.

Goal of this lesson: understand why complementary-angle ratios are connected, learn the identities, and use them to simplify expressions and solve problems.

1. Quick review: trigonometric ratios in a right triangle

Take a right triangle and choose one acute angle, say \(\theta\). The sides are named relative to that angle:

• Opposite: the side across from \(\theta\)
• Adjacent: the side next to \(\theta\) (not the hypotenuse)
• Hypotenuse: the longest side, opposite the right angle

The main trigonometric ratios are:

There are also three related ratios:

2. Why complementary angles switch roles

Suppose a right triangle has acute angles \(\theta\) and \(90^\circ - \theta\).

Now focus on the side names. A very important thing happens:

• The side that is opposite to \(\theta\) becomes adjacent to \(90^\circ-\theta\).
• The side that is adjacent to \(\theta\) becomes opposite to \(90^\circ-\theta\).
• The hypotenuse stays the same.

This means sine and cosine swap, tangent and cotangent swap, and secant and cosecant swap.

3. The complementary angle identities

These are the identities you should know:

These identities are true for acute angles in right-triangle trigonometry.

4. Understanding one identity clearly

Let one acute angle in a right triangle be \(x\). Then the other acute angle is \(90^\circ-x\).

For angle \(x\):

For angle \(90^\circ-x\), that same opposite side to \(x\) is now the adjacent side. So:

Therefore,

Using the same idea, all the other identities can be understood.

5. A helpful memory idea

When the angle changes from \(x\) to \(90^\circ-x\), the trigonometric ratio changes to its co-function:

• \(\sin \leftrightarrow \cos\)
• \(\tan \leftrightarrow \cot\)
• \(\sec \leftrightarrow \csc\)

So, if you see \(90^\circ-x\), think: switch to the matching co-function.

6. Worked Example 1: Basic identity use

Simplify \(\sin 58^\circ\).

Notice that \(58^\circ = 90^\circ - 32^\circ\). So:

Using \(\sin(90^\circ-x)=\cos x\):

Therefore,

Answer: \(\cos 32^\circ\)

7. Worked Example 2: Using tangent and cotangent

Simplify \(\tan 25^\circ\) in terms of cotangent.

Since \(25^\circ = 90^\circ - 65^\circ\), we write:

Using \(\tan(90^\circ-x)=\cot x\):

So,

Answer: \(\cot 65^\circ\)

8. Worked Example 3: Simplifying an expression

Simplify:

Use the identity \(\sin(90^\circ-x)=\cos x\).

Then the expression becomes:

Answer: \(1\)

This kind of question is very common. The key step is replacing the complementary-angle ratio with its matching ratio.

9. Worked Example 4: A slightly harder expression

Simplify:

Use the identity:

So the expression becomes:

Now use \(\csc \theta = \frac{1}{\sin \theta}\):

Answer: \(1\)

10. How to solve these questions step by step

1. Look for an angle of the form \(90^\circ-x\).
2. Identify the trigonometric ratio attached to it.
3. Replace it with the matching co-function:
   • \(\sin \to \cos\)
   • \(\cos \to \sin\)
   • \(\tan \to \cot\)
   • \(\cot \to \tan\)
   • \(\sec \to \csc\)
   • \(\csc \to \sec\)
4. Simplify the new expression.

11. Common mistakes to avoid

• Do not keep the same ratio. For example, \(\sin(90^\circ-x)\) is not \(\sin x\). It becomes \(\cos x\).
• Watch the angle carefully. These identities are for complementary angles, so the angle must be of the form \(90^\circ-x\).
• Do not mix up reciprocal and complementary ideas. For example, \(\sin x\) and \(\csc x\) are reciprocals, but \(\sin(90^\circ-x)\) becomes \(\cos x\), not \(\csc x\).
• Check whether the question asks to simplify or evaluate. Sometimes you only need to rewrite the expression, not find a decimal value.

12. Quick practice ideas

Try rewriting these on your own:

• \(\cos 17^\circ = \sin 73^\circ\)
• \(\csc 40^\circ = \sec 50^\circ\)
• \(\cot 12^\circ = \tan 78^\circ\)
• \(\frac{\tan(90^\circ-x)}{\cot x}\)

For the last one, use \(\tan(90^\circ-x)=\cot x\), so the fraction becomes \(\frac{\cot x}{\cot x}=1\).

13. Summary

Complementary angles are two angles that add up to \(90^\circ\). In a right triangle, the two acute angles are always complementary.

Because the opposite and adjacent sides switch roles when you move from one acute angle to the other, the trigonometric ratios also switch in pairs:

These identities are very useful for simplifying expressions quickly and correctly. Whenever you see \(90^\circ-x\), think: switch to the matching co-function.

Introduction

In trigonometry, identities are equations that are always true for allowed values of the variable. One of the most important identities is the Fundamental Pythagorean Identity:

$$\sin^2\theta + \cos^2\theta = 1$$

This means that for any angle \(\theta\), the square of the sine plus the square of the cosine will always equal 1.

This identity is called “Pythagorean” because it comes directly from the Pythagorean Theorem, which you already know from right triangles:

$$a^2+b^2=c^2$$

In this lesson, you will learn what the identity means, why it is true, and how to use it to find missing trig values.

Review: Sine and Cosine in a Right Triangle

Suppose \(\theta\) is one acute angle in a right triangle.

• \(\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}\)
• \(\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}\)

If we label the opposite side as \(o\), the adjacent side as \(a\), and the hypotenuse as \(h\), then:

$$\sin\theta = \frac{o}{h} \qquad \text{and} \qquad \cos\theta = \frac{a}{h}$$

Where the Identity Comes From

Start with the Pythagorean Theorem for a right triangle:

$$o^2+a^2=h^2$$

Now divide every term by \(h^2\):

$$\frac{o^2}{h^2}+\frac{a^2}{h^2}=\frac{h^2}{h^2}$$

Simplify each fraction:

$$\left(\frac{o}{h}\right)^2+\left(\frac{a}{h}\right)^2=1$$

Replace the ratios with sine and cosine:

$$\sin^2\theta+\cos^2\theta=1$$

This proves the Fundamental Pythagorean Identity.

What Does \(\sin^2\theta\) Mean?

The notation \(\sin^2\theta\) means:

$$\sin^2\theta = (\sin\theta)^2$$

It does not mean \(\sin(\theta^2)\).

In the same way:

$$\cos^2\theta = (\cos\theta)^2$$

Why This Identity Matters

This identity is useful because if you know the sine or cosine of an angle, you can find the other one.

Starting with

$$\sin^2\theta+\cos^2\theta=1$$

you can rearrange it in two helpful ways:

$$\sin^2\theta = 1-\cos^2\theta$$

$$\cos^2\theta = 1-\sin^2\theta$$

Then take the square root to find the missing value.

Be careful: when taking a square root, there can be a positive or negative answer. In right triangle problems with acute angles, sine and cosine are positive. In other settings, you may need to think about the sign.

Worked Example 1: Verify the Identity Using a Right Triangle

A right triangle has side lengths 3, 4, and 5. Let \(\theta\) be the angle opposite the side of length 3.

First find sine and cosine:

$$\sin\theta = \frac{3}{5} \qquad \cos\theta = \frac{4}{5}$$

Now square each value:

$$\sin^2\theta = \left(\frac{3}{5}\right)^2 = \frac{9}{25}$$

$$\cos^2\theta = \left(\frac{4}{5}\right)^2 = \frac{16}{25}$$

Add them:

$$\sin^2\theta + \cos^2\theta = \frac{9}{25}+\frac{16}{25}=\frac{25}{25}=1$$

So the identity is true for this triangle.

Worked Example 2: Find Cosine from a Given Sine

If \(\sin\theta=\frac{5}{13}\) and \(\theta\) is an acute angle, find \(\cos\theta\).

Use the identity:

$$\sin^2\theta+\cos^2\theta=1$$

Substitute \(\sin\theta=\frac{5}{13}\):

$$\left(\frac{5}{13}\right)^2+\cos^2\theta=1$$

$$\frac{25}{169}+\cos^2\theta=1$$

Subtract \(\frac{25}{169}\) from both sides:

$$\cos^2\theta=1-\frac{25}{169}$$

Write 1 as \(\frac{169}{169}\):

$$\cos^2\theta=\frac{169}{169}-\frac{25}{169}=\frac{144}{169}$$

Take the square root:

$$\cos\theta=\pm\frac{12}{13}$$

Because \(\theta\) is acute in a right triangle, cosine is positive.

$$\cos\theta=\frac{12}{13}$$

Worked Example 3: Find Sine from a Given Cosine

If \(\cos\theta=0.8\) and \(\theta\) is acute, find \(\sin\theta\).

Use the identity:

$$\sin^2\theta+\cos^2\theta=1$$

Substitute \(\cos\theta=0.8\):

$$\sin^2\theta+(0.8)^2=1$$

$$\sin^2\theta+0.64=1$$

Subtract 0.64:

$$\sin^2\theta=0.36$$

Take the square root:

$$\sin\theta=\pm 0.6$$

Since \(\theta\) is acute, sine is positive.

$$\sin\theta=0.6$$

Worked Example 4: Use the Identity in Algebra

Simplify:

$$\sin^2\theta + \cos^2\theta + 7$$

Use the identity \(\sin^2\theta + \cos^2\theta = 1\):

$$1+7=8$$

So the simplified expression is:

$$8$$

Common Mistakes to Avoid

• Mixing up the ratios: Remember that sine uses opposite over hypotenuse, and cosine uses adjacent over hypotenuse.
• Misreading exponents: \(\sin^2\theta\) means \((\sin\theta)^2\), not \(\sin(\theta^2)\).
• Forgetting to square first: In the identity, the sine and cosine values are squared before being added.
• Ignoring the sign after square roots: If the problem says the angle is acute, use the positive value.

Key Idea to Remember

The identity works because sine and cosine come from side ratios in a right triangle, and the side lengths of a right triangle always satisfy the Pythagorean Theorem.

No matter which angle you choose,

$$\sin^2\theta + \cos^2\theta = 1$$

Quick Practice

1. If \(\sin\theta=\frac{8}{17}\) and \(\theta\) is acute, find \(\cos\theta\).
2. If \(\cos\theta=\frac{15}{17}\) and \(\theta\) is acute, find \(\sin\theta\).
3. Simplify: \(3(\sin^2\theta+\cos^2\theta)\).

Answers

1. $$\cos\theta=\frac{15}{17}$$
2. $$\sin\theta=\frac{8}{17}$$
3. $$3(1)=3$$

Summary

The Fundamental Pythagorean Identity is

$$\sin^2\theta+\cos^2\theta=1$$

It comes from the Pythagorean Theorem and the definitions of sine and cosine in a right triangle.

You can use this identity to verify trig values, simplify expressions, and find a missing sine or cosine when you know the other one.

Quotient and Reciprocal Identities are important trigonometric relationships that help us rewrite expressions in simpler forms. These identities connect the six trig functions: sine, cosine, tangent, cotangent, secant, and cosecant.

In this lesson, you will learn how tangent and cotangent can be written as quotients, and how sine, cosine, and tangent each have reciprocal partners. These relationships make it easier to simplify expressions, solve equations, and understand how trig functions are connected.

Before starting, remember the three basic trig ratios in a right triangle:

• $$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$$
• $$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$$
• $$\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$$

These basic definitions lead directly to the quotient identities and reciprocal identities.

1. Quotient Identities

The quotient identities show how tangent and cotangent can be written using sine and cosine.

• $$\tan \theta = \frac{\sin \theta}{\cos \theta}$$
• $$\cot \theta = \frac{\cos \theta}{\sin \theta}$$

Why does this work? Start with the definitions from a right triangle:

$$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}, \qquad \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$$

Now divide sine by cosine:

$$\frac{\sin \theta}{\cos \theta} = \frac{\frac{\text{opposite}}{\text{hypotenuse}}}{\frac{\text{adjacent}}{\text{hypotenuse}}}$$

When dividing fractions, multiply by the reciprocal:

$$\frac{\text{opposite}}{\text{hypotenuse}} \cdot \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{\text{opposite}}{\text{adjacent}} = \tan \theta$$

So, $$\tan \theta = \frac{\sin \theta}{\cos \theta}$$.

In the same way, dividing cosine by sine gives:

$$\frac{\cos \theta}{\sin \theta} = \frac{\text{adjacent}}{\text{opposite}} = \cot \theta$$

So, $$\cot \theta = \frac{\cos \theta}{\sin \theta}$$.

2. Reciprocal Identities

Reciprocal identities come from pairs of trig functions that are opposites in multiplication. Two numbers are reciprocals if their product is 1.

The reciprocal identities are:

• $$\csc \theta = \frac{1}{\sin \theta} \qquad \text{and} \qquad \sin \theta = \frac{1}{\csc \theta}$$
• $$\sec \theta = \frac{1}{\cos \theta} \qquad \text{and} \qquad \cos \theta = \frac{1}{\sec \theta}$$
• $$\cot \theta = \frac{1}{\tan \theta} \qquad \text{and} \qquad \tan \theta = \frac{1}{\cot \theta}$$

This means:

• sine and cosecant are reciprocals,
• cosine and secant are reciprocals,
• tangent and cotangent are reciprocals.

For example, if $$\sin \theta = \frac{3}{5}$$, then:

$$\csc \theta = \frac{1}{\sin \theta} = \frac{1}{\frac{3}{5}} = \frac{5}{3}$$

3. Why These Identities Matter

Quotient and reciprocal identities help you:

• rewrite expressions in different trig forms,
• simplify complicated fractions,
• solve trigonometric equations,
• check whether two expressions are equivalent.

Many trig problems become easier when you change everything into sine and cosine first.

4. Worked Examples

Example 1: Rewrite tangent using sine and cosine

Rewrite $$\tan x$$ as a quotient.

Use the quotient identity:

$$\tan x = \frac{\sin x}{\cos x}$$

Answer: $$\frac{\sin x}{\cos x}$$

Example 2: Find a reciprocal function

If $$\cos \theta = \frac{4}{7}$$, find $$\sec \theta$$.

Since secant is the reciprocal of cosine:

$$\sec \theta = \frac{1}{\cos \theta} = \frac{1}{\frac{4}{7}} = \frac{7}{4}$$

Answer: $$\sec \theta = \frac{7}{4}$$

Example 3: Simplify using a quotient identity

Simplify:

$$\frac{\sin x}{\cos x}$$

By the quotient identity,

$$\frac{\sin x}{\cos x} = \tan x$$

Answer: $$\tan x$$

Example 4: Simplify a more complex expression

Simplify:

$$\frac{1}{\csc x} \cdot \frac{\cos x}{\sin x}$$

Step 1: Use the reciprocal identity $$\frac{1}{\csc x} = \sin x$$.

So the expression becomes:

$$\sin x \cdot \frac{\cos x}{\sin x}$$

Step 2: Cancel $$\sin x$$ from top and bottom:

$$\sin x \cdot \frac{\cos x}{\sin x} = \cos x$$

Answer: $$\cos x$$

5. Common Mistakes to Avoid

• Mixing up quotient and reciprocal identities: $$\tan \theta = \frac{\sin \theta}{\cos \theta}$$, not $$\frac{1}{\sin \theta}$$.
• Forgetting which functions are pairs: sine pairs with cosecant, cosine pairs with secant, tangent pairs with cotangent.
• Dividing incorrectly: when dividing fractions, multiply by the reciprocal.
• Cancelling too early: only cancel factors, not terms that are being added or subtracted.

6. Quick Identity List to Memorize

• $$\tan \theta = \frac{\sin \theta}{\cos \theta}$$
• $$\cot \theta = \frac{\cos \theta}{\sin \theta}$$
• $$\csc \theta = \frac{1}{\sin \theta}$$
• $$\sec \theta = \frac{1}{\cos \theta}$$
• $$\cot \theta = \frac{1}{\tan \theta}$$

7. Final Summary

The quotient identities show that tangent and cotangent can be written as fractions of sine and cosine:

$$\tan \theta = \frac{\sin \theta}{\cos \theta}, \qquad \cot \theta = \frac{\cos \theta}{\sin \theta}$$

The reciprocal identities show that sine, cosine, and tangent each have a reciprocal partner:

$$\csc \theta = \frac{1}{\sin \theta}, \qquad \sec \theta = \frac{1}{\cos \theta}, \qquad \cot \theta = \frac{1}{\tan \theta}$$

If you can recognize these identities quickly, you will be able to simplify trig expressions much more easily. A good strategy is to rewrite everything in terms of sine and cosine whenever you feel stuck.

Proving Complex Trigonometric Identities

In trigonometry, an identity is an equation that is true for all values of the variable where both sides are defined. For example, the identity

$$\sin^2\theta + \cos^2\theta = 1$$

is always true, no matter what angle \(\theta\) is.

When you are asked to prove a trigonometric identity, your goal is not to plug in numbers. Instead, you use known trig identities and algebra skills to change one side of the equation until it matches the other side.

This lesson will show you how to prove more complicated identities step by step using the most important trig facts, careful algebra, and good strategy.

1. What does it mean to prove an identity?

Suppose you are given:

$$\frac{1-\cos^2\theta}{\sin\theta} = \sin\theta$$

To prove this identity, you want to show that the left side simplifies to the right side, or that the right side simplifies to the left side.

You are not solving for \(\theta\). You are showing that both expressions are really the same expression written in different forms.

2. Important identities you must know

Most proofs at this level use a small group of basic identities. Memorizing them will make proofs much easier.

• Pythagorean identity: $$\sin^2\theta + \cos^2\theta = 1$$
• From that identity: $$1-\sin^2\theta = \cos^2\theta$$
• Also: $$1-\cos^2\theta = \sin^2\theta$$
• Tangent identity: $$\tan\theta = \frac{\sin\theta}{\cos\theta}$$

These are the tools you will use again and again.

3. Key idea: work on one side only

A common mistake is to simplify both sides at the same time. This can get confusing and may not actually prove anything.

Usually, the best strategy is to start with the side that looks more complicated and simplify only that side until it becomes the other side.

4. Useful algebra skills in trig proofs

Proving trig identities is really a mix of trigonometry and algebra. These algebra skills are especially important:

• Factoring
• Finding common denominators
• Splitting fractions carefully
• Substituting identities like \(1-\sin^2\theta = \cos^2\theta\)
• Rewriting \(\tan\theta\) as \(\frac{\sin\theta}{\cos\theta}\)
• Canceling only common factors, not terms

5. A step-by-step strategy for proving identities

1. Look at both sides and decide which side seems more complicated.
2. Rewrite everything in terms of \(\sin\theta\) and \(\cos\theta\) if needed.
3. Use the Pythagorean identity when you see expressions like \(1-\sin^2\theta\) or \(1-\cos^2\theta\).
4. Factor if possible.
5. Combine fractions or split them when helpful.
6. Stop as soon as one side matches the other.

6. Worked Example 1: Using the Pythagorean identity

Prove:

$$\frac{1-\cos^2\theta}{\sin\theta} = \sin\theta$$

Step 1: Start with the left side because it looks more complicated.

$$\frac{1-\cos^2\theta}{\sin\theta}$$

Step 2: Use the identity \(1-\cos^2\theta = \sin^2\theta\).

$$\frac{\sin^2\theta}{\sin\theta}$$

Step 3: Simplify by canceling the common factor \(\sin\theta\).

$$\sin\theta$$

This matches the right side, so the identity is proved.

7. Worked Example 2: Rewriting tangent

Prove:

$$\frac{\tan\theta}{\sin\theta} = \frac{1}{\cos\theta}$$

Step 1: Start with the left side.

$$\frac{\tan\theta}{\sin\theta}$$

Step 2: Rewrite \(\tan\theta\) as \(\frac{\sin\theta}{\cos\theta}\).

$$\frac{\frac{\sin\theta}{\cos\theta}}{\sin\theta}$$

Step 3: Dividing by \(\sin\theta\) means multiplying by \(\frac{1}{\sin\theta}\).

$$\frac{\sin\theta}{\cos\theta} \cdot \frac{1}{\sin\theta}$$

Step 4: Cancel \(\sin\theta\).

$$\frac{1}{\cos\theta}$$

This matches the right side, so the identity is proved.

8. Worked Example 3: Factoring and simplifying

Prove:

$$\frac{\sin^2\theta-1}{\cos\theta} = -\cos\theta$$

Step 1: Start with the left side.

$$\frac{\sin^2\theta-1}{\cos\theta}$$

Step 2: Notice that

$$\sin^2\theta - 1 = -(1-\sin^2\theta)$$

Using \(1-\sin^2\theta = \cos^2\theta\), we get

$$\sin^2\theta - 1 = -\cos^2\theta$$

So the expression becomes

$$\frac{-\cos^2\theta}{\cos\theta}$$

Step 3: Simplify.

$$-\cos\theta$$

This matches the right side, so the identity is proved.

9. Worked Example 4: Combining fractions

Prove:

$$\frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\cos\theta} = \frac{\sin\theta + \cos\theta}{\cos\theta}$$

This example is more algebraic, but it is still important in trig proofs.

Step 1: Start with the left side.

$$\frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\cos\theta}$$

Step 2: Since the denominators are the same, combine the numerators.

$$\frac{\sin\theta + \cos\theta}{\cos\theta}$$

This matches the right side, so the identity is proved.

10. Common mistakes to avoid

• Do not treat an identity like an equation to solve. You are simplifying, not finding values of \(\theta\).
• Do not simplify both sides randomly. Usually, simplify only one side.
• Do not cancel terms that are added or subtracted. For example, in $$\frac{\sin\theta+\cos\theta}{\cos\theta}$$ you cannot cancel the \(\cos\theta\) from just one term.
• Use identities correctly. For example, \(1-\sin^2\theta = \cos^2\theta\), not \(\cos\theta\).
• Be careful with negative signs. These often cause errors in proofs.

11. How to decide what to do first

Sometimes students get stuck because they do not know which identity to use. Here are some helpful clues:

• If you see \(\tan\theta\), try rewriting it as \(\frac{\sin\theta}{\cos\theta}\).
• If you see \(1-\sin^2\theta\), replace it with \(\cos^2\theta\).
• If you see \(1-\cos^2\theta\), replace it with \(\sin^2\theta\).
• If you see fractions, look for a common denominator or possible factoring.
• If one side is very simple, try turning the more complex side into that simple form.

12. Quick practice ideas

Try proving these on your own using the same methods:

• $$\frac{1-\sin^2\theta}{\cos\theta} = \cos\theta$$
• $$\tan\theta \cdot \cos\theta = \sin\theta$$
• $$\frac{\sin^2\theta}{\cos\theta} + \cos\theta = \frac{1}{\cos\theta}$$

For the last one, try combining the terms on the left using a common denominator. Then use \(\sin^2\theta + \cos^2\theta = 1\).

13. Final summary

To prove complex trigonometric identities, start with one side of the equation and use basic trig identities plus algebra to transform it into the other side.

The most important tools are:

• $$\sin^2\theta + \cos^2\theta = 1$$
• $$1-\sin^2\theta = \cos^2\theta$$
• $$1-\cos^2\theta = \sin^2\theta$$
• $$\tan\theta = \frac{\sin\theta}{\cos\theta}$$

As you practice, you will get better at noticing patterns and choosing the right identity. The key is to work carefully, show each step clearly, and use algebra correctly.

Angles of Elevation and Depression are used in trigonometry to describe how we look upward or downward at an object.

These angles help us turn real-life situations into right triangles so that we can use trigonometric ratios like sine, cosine, and tangent.

This lesson will show you what these angles mean, how to draw the correct diagram, and how to solve problems involving heights and distances.

1. What is an angle of elevation?

An angle of elevation is the angle formed when you look up at an object from a horizontal line.

Imagine standing on the ground and looking at the top of a tree, building, or flagpole. The angle between your straight-ahead horizontal line and your line of sight upward is the angle of elevation.

2. What is an angle of depression?

An angle of depression is the angle formed when you look down at an object from a horizontal line.

Imagine standing on the top of a lighthouse and looking down at a boat in the water. The angle between your horizontal line and your line of sight downward is the angle of depression.

Important idea: both angles are always measured from a horizontal line, not from a vertical line.

3. Line of sight

The line of sight is the straight line from the observer's eyes to the object being viewed.

In most problems, the line of sight becomes the hypotenuse of a right triangle.

4. Drawing the diagram correctly

When solving these problems, drawing the right triangle is the most important step.

• Draw a horizontal line from the observer.
• Draw the line of sight to the object.
• Mark the given angle from the horizontal.
• Add the vertical and horizontal distances to complete a right triangle.

Usually:

• The vertical side represents a height difference.
• The horizontal side represents the ground distance.
• The slanted side represents the line of sight.

5. Why are angles of elevation and depression related?

In many problems, an angle of depression from one point is equal to the angle of elevation from the other point.

This happens because the two horizontal lines are parallel, and the line of sight acts like a transversal. The two angles formed are equal alternate interior angles.

So if a person on a cliff looks down at a boat with an angle of depression of \(35^\circ\), then the boat sees the top of the cliff at an angle of elevation of \(35^\circ\).

6. Choosing the correct trigonometric ratio

After drawing the triangle, identify which sides you know and which side you need.

Use:

• \(\sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}}\)
• \(\cos \theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}\)
• \(\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}}\)

In many elevation and depression problems, we use tangent because the problem often gives a height and a horizontal distance.

That means:

7. Steps for solving word problems

1. Read the problem carefully.
2. Draw and label a right triangle.
3. Decide whether the angle is one of elevation or depression.
4. Mark the angle from a horizontal line.
5. Identify the opposite, adjacent, and hypotenuse sides.
6. Choose the correct trig ratio.
7. Solve the equation.
8. Check whether your answer makes sense.

Worked Example 1: Finding height from distance and angle

A student stands \(20\) m from the base of a tree. The angle of elevation to the top of the tree is \(32^\circ\). Find the height of the tree.

Step 1: Identify the sides.

The horizontal distance is \(20\) m, so this is the adjacent side.

The height of the tree is the opposite side.

Step 2: Choose the ratio.

Since we have opposite and adjacent, use tangent:

Step 3: Solve.

Answer: The tree is about \(12.5\) m tall.

Worked Example 2: Finding horizontal distance from height and angle

A kite is flying at a height of \(18\) m above the ground. The angle of elevation from a point on the ground to the kite is \(40^\circ\). How far is the point on the ground from the point directly below the kite?

Step 1: Identify the sides.

The height \(18\) m is the opposite side.

The horizontal distance \(x\) is the adjacent side.

Step 2: Use tangent.

Step 3: Solve for \(x\).

Answer: The horizontal distance is about \(21.5\) m.

Worked Example 3: Using angle of depression

A person stands on a balcony that is \(25\) m above the ground. The angle of depression to a car on the road is \(28^\circ\). How far is the car from the building?

Step 1: Use the angle relationship.

The angle of depression from the balcony equals the angle of elevation from the car, so the angle in the triangle is also \(28^\circ\).

Step 2: Identify the sides.

The height \(25\) m is the opposite side.

The horizontal distance \(d\) is the adjacent side.

Step 3: Use tangent.

Step 4: Solve.

Answer: The car is about \(47.0\) m from the building.

Worked Example 4: Finding the line of sight

A ladder leans against a wall. The angle of elevation from the ground to the top of the ladder is \(65^\circ\), and the base of the ladder is \(4\) m from the wall. Find the length of the ladder.

Step 1: Identify the sides.

The horizontal distance \(4\) m is the adjacent side.

The ladder length \(L\) is the hypotenuse.

Step 2: Choose cosine.

Step 3: Solve.

Answer: The ladder is about \(9.5\) m long.

8. Common mistakes to avoid

• Measuring from the wrong line: Always measure elevation or depression from a horizontal line.
• Mixing up opposite and adjacent: These depend on where the angle is located.
• Using the wrong trig ratio: Choose sine, cosine, or tangent based on the sides involved.
• Forgetting equal angles: The angle of depression often equals the angle of elevation.
• Not checking units: Make sure your final answer includes meters, feet, or whatever unit is used.

9. Quick check for understanding

Ask yourself these questions:

• Am I looking up or looking down?
• Did I draw the horizontal line correctly?
• Which side is opposite the angle?
• Which side is adjacent?
• Should I use sine, cosine, or tangent?

10. Summary

An angle of elevation is the angle formed when looking upward from a horizontal line, and an angle of depression is the angle formed when looking downward from a horizontal line.

These situations can be modeled using right triangles. Once the triangle is drawn correctly, trigonometric ratios help find unknown heights, distances, or line-of-sight lengths.

The most important skills are drawing the diagram carefully, measuring angles from the horizontal, and choosing the correct trigonometric ratio.

Solving Single-Triangle Height and Distance Problems

In many real-life situations, we cannot measure a height directly. For example, it may be difficult to measure the height of a tree, a building, or a flagpole by climbing it. Trigonometry helps us find these unknown heights using a right triangle, one known side length, and one known angle.

In this lesson, you will learn how to solve single-triangle height and distance problems. These are problems where one right triangle is enough to model the situation. You will use the trigonometric ratios sine, cosine, and tangent to find missing lengths.

1. Understanding the right triangle in a height-and-distance problem

Most height-and-distance problems involve an observer looking at the top of an object. This creates a right triangle:

• The height of the object is usually the vertical side.
• The distance from the observer to the object is usually the horizontal side.
• The line of sight from the observer to the top of the object is the hypotenuse.

The angle used is often called the angle of elevation.

• Angle of elevation: the angle measured upward from a horizontal line to an object above eye level.
• Angle of depression: the angle measured downward from a horizontal line to an object below eye level.

For this lesson, we will mainly focus on angle of elevation problems.

2. Reviewing the trigonometric ratios

In a right triangle, the three main trigonometric ratios are:

Here, opposite means the side across from the angle, and adjacent means the side next to the angle that is not the hypotenuse.

To choose the correct ratio, look at which sides you know and which side you need to find.

• Use sine when you have or need the opposite and hypotenuse.
• Use cosine when you have or need the adjacent and hypotenuse.
• Use tangent when you have or need the opposite and adjacent.

3. Why tangent is often used in height problems

In many height-and-distance questions, you are given:

• the horizontal distance from the object, and
• the angle of elevation.

You are asked to find the height. In this case, the height is the opposite side, and the horizontal distance is the adjacent side. That means tangent is usually the best choice:

Then solve for the height:

4. A step-by-step method

1. Draw a right triangle to match the situation.
2. Label the known angle and the known side.
3. Identify the unknown side.
4. Choose the correct trig ratio using opposite, adjacent, and hypotenuse.
5. Write an equation.
6. Solve carefully.
7. Check if the answer makes sense in the real situation.

5. Important detail: eye level

Sometimes the person measuring the angle is not standing on the ground level of the triangle’s bottom corner. For example, the measuring tool might be held at eye level, such as 1.6 m above the ground.

In that case, trigonometry first gives you the height above the observer’s eyes, not the full height of the object.

So if needed:

Worked Example 1: Finding the height of a tree using tangent

A student stands 12 m from a tree. The angle of elevation to the top of the tree is \(38^\circ\). Find the height of the tree.

Step 1: Identify the sides

• Unknown height of tree = opposite side
• Distance from student to tree = 12 m = adjacent side
• Angle = \(38^\circ\)

Step 2: Choose a trig ratio

We need opposite and adjacent, so use tangent:

Step 3: Solve

The height of the tree is about \(9.4\text{ m}\).

Worked Example 2: Finding the distance from a building using tangent

A person looks at the top of a building with an angle of elevation of \(52^\circ\). The building is 20 m tall. How far is the person standing from the building?

Step 1: Identify the sides

• Height of building = 20 m = opposite side
• Unknown horizontal distance = adjacent side
• Angle = \(52^\circ\)

Step 2: Use tangent

Step 3: Solve for \(d\)

The person is standing about \(15.6\text{ m}\) from the building.

Worked Example 3: Using sine when the hypotenuse is known

A kite string is 25 m long and makes an angle of \(41^\circ\) with the ground. Assuming the string is straight and tight, how high is the kite above the ground?

Step 1: Identify the sides

• String length = 25 m = hypotenuse
• Height of kite = opposite side
• Angle with ground = \(41^\circ\)

Step 2: Choose sine

Step 3: Solve

The kite is about \(16.4\text{ m}\) above the ground.

Worked Example 4: Including the observer’s eye level

A surveyor stands 30 m from a tower. The angle of elevation from the surveyor’s eye to the top of the tower is \(47^\circ\). The surveyor’s eye is 1.5 m above the ground. Find the total height of the tower.

Step 1: Set up the triangle

• Horizontal distance = 30 m = adjacent side
• Height from eye level to top of tower = opposite side
• Angle = \(47^\circ\)

Step 2: Use tangent

Step 3: Solve for \(h\)

This \(32.17\text{ m}\) is the height from the surveyor’s eyes to the top of the tower.

Step 4: Add the eye level

The tower is about \(33.7\text{ m}\) tall.

6. How to decide which trig ratio to use

Ask yourself these questions:

• What angle am I using?
• Which side is opposite that angle?
• Which side is adjacent?
• Is the hypotenuse involved?

A quick guide:

• If you see height and distance, think tangent.
• If you see height and slanted side, think sine.
• If you see distance and slanted side, think cosine.

7. Common mistakes to avoid

• Using the wrong angle: Be sure you use the angle given in the triangle.
• Mixing up opposite and adjacent: Always identify them from the given angle.
• Forgetting eye level: Add the observer’s height if the angle is measured from eye level.
• Calculator in the wrong mode: Make sure your calculator is in degrees, not radians.
• Rounding too early: Keep extra decimal places until the final answer.

8. Practice thinking

Here is how you should think through a problem:

“I know the angle and the ground distance. I want the height. Height is opposite, distance is adjacent, so I should use tangent.”

Or:

“I know the angle and the slanted side. I want the height. Height is opposite, slanted side is the hypotenuse, so I should use sine.”

9. Final summary

Single-triangle height and distance problems use one right triangle to model a real situation. You identify the angle, label the sides as opposite, adjacent, and hypotenuse, and then choose sine, cosine, or tangent to find the missing length.

Most height problems use tangent because height is often opposite the angle and ground distance is adjacent. Always draw a diagram, label carefully, and remember to include eye level if the problem mentions it.

Solving Multi-Triangle Height and Distance Problems

In trigonometry, height and distance problems often involve a right triangle, an angle, and a side you need to find. A multi-triangle problem is more challenging because it uses two related triangles in the same situation.

These problems usually involve:

• Two different angles of elevation to the same object, or
• An observer moving closer to or farther from an object, creating a second triangle.

The key idea is that the triangles are connected by the same height, the same ground line, or a known change in distance. If you can describe those relationships carefully, you can set up equations and solve the problem step by step.

1. Review: the main trigonometric ratios

In a right triangle:

• $$\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$$
• $$\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}$$
• $$\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}$$

For most height and distance problems, tangent is the most useful ratio because it connects height and horizontal distance directly:

$$\tan(\theta)=\frac{\text{height}}{\text{distance}}$$

2. Important vocabulary

• Angle of elevation: the angle measured upward from a horizontal line of sight.
• Horizontal distance: the distance along the ground from the observer to the base of the object.
• Height: the vertical side of the triangle.

If the observer is standing on level ground and looking at the top of a building, tree, tower, or cliff, then the ground and the object usually form a right triangle.

3. Why multi-triangle problems are different

In a single-triangle problem, one trig equation is often enough. In a multi-triangle problem, you often have two unknowns, such as:

• the height of the object, and
• the distance from the observer to the object.

Because there are two unknowns, you usually need two equations. The second triangle gives you that second equation.

4. Common setup strategy

When solving these problems, use this method:

1. Draw or imagine the diagram clearly.
2. Label the height with a variable such as \(h\).
3. Label the unknown distance with a variable such as \(x\).
4. Use the second triangle to write another expression, often \(x+\text{something}\) or \(x-\text{something}\).
5. Write a tangent equation for each triangle.
6. Solve the system of equations.
7. Check whether the answer makes sense.

5. Pattern A: Two angles of elevation from two points on the same line

Suppose one observer is farther away and another is closer. Both look at the top of the same building. The closer observer should have the larger angle of elevation, because the building appears steeper from closer up.

If the closer distance is \(x\), and the farther point is 20 m farther away, then the farther distance is \(x+20\).

If the height is \(h\), then the two equations are:

$$\tan(\theta_1)=\frac{h}{x}$$

$$\tan(\theta_2)=\frac{h}{x+20}$$

Since both equal the same height expression, you can solve for \(x\) and then find \(h\).

Worked Example 1: Finding the height of a building

A student stands some distance from a building and measures the angle of elevation to the top as \(50^\circ\). The student then walks 30 m farther away, and the angle becomes \(35^\circ\). Find the height of the building.

Step 1: Define variables

• Let \(x\) be the original distance from the building.
• Then the new distance is \(x+30\).
• Let \(h\) be the height of the building.

Step 2: Write tangent equations

From the first position:

$$\tan(50^\circ)=\frac{h}{x}$$

So:

$$h=x\tan(50^\circ)$$

From the second position:

$$\tan(35^\circ)=\frac{h}{x+30}$$

So:

$$h=(x+30)\tan(35^\circ)$$

Step 3: Set the two height expressions equal

$$x\tan(50^\circ)=(x+30)\tan(35^\circ)$$

Using decimal values:

$$x(1.1918)=(x+30)(0.7002)$$

Expand:

$$1.1918x=0.7002x+21.006$$

Subtract \(0.7002x\) from both sides:

$$0.4916x=21.006$$

$$x\approx 42.73$$

Step 4: Find the height

$$h=x\tan(50^\circ)$$

$$h\approx 42.73(1.1918)\approx 50.93$$

Answer: The building is about \(50.9\) m tall.

6. Pattern B: Two observers at different distances at the same time

Sometimes two people observe the same object from two different points. If the distance between the observers is known, the setup is very similar.

The main question is: Which distance is longer? The smaller angle usually belongs to the observer who is farther away.

Worked Example 2: Two observers looking at a tower

Two observers stand on the same straight path from a tower. The closer observer measures an angle of elevation of \(60^\circ\), and the farther observer measures \(40^\circ\). They are 25 m apart. Find the height of the tower.

Step 1: Define variables

• Let \(x\) be the distance from the closer observer to the tower.
• Then the farther observer is \(x+25\) m from the tower.
• Let \(h\) be the height of the tower.

Step 2: Write equations

$$\tan(60^\circ)=\frac{h}{x} \Rightarrow h=x\tan(60^\circ)$$

$$\tan(40^\circ)=\frac{h}{x+25} \Rightarrow h=(x+25)\tan(40^\circ)$$

Step 3: Set them equal

$$x\tan(60^\circ)=(x+25)\tan(40^\circ)$$

Substitute decimal values:

$$x(1.7321)=(x+25)(0.8391)$$

$$1.7321x=0.8391x+20.9775$$

$$0.893x=20.9775$$

$$x\approx 23.49$$

Step 4: Find height

$$h=x\tan(60^\circ)$$

$$h\approx 23.49(1.7321)\approx 40.69$$

Answer: The tower is about \(40.7\) m tall.

7. Pattern C: The observer moves toward the object

If the observer moves closer to the object, the distance gets smaller. So if the original distance is \(x\), and the observer moves 15 m closer, the new distance is \(x-15\).

This is an important place where students make mistakes. Always check whether the movement should make the distance bigger or smaller.

Worked Example 3: Moving closer to a tree

A person is standing some distance from a tree. The angle of elevation to the top is \(28^\circ\). After walking 18 m closer, the angle becomes \(43^\circ\). Find the height of the tree.

Step 1: Define variables

• Let \(x\) be the original distance from the tree.
• After walking closer, the new distance is \(x-18\).
• Let \(h\) be the height of the tree.

Step 2: Write equations

$$\tan(28^\circ)=\frac{h}{x} \Rightarrow h=x\tan(28^\circ)$$

$$\tan(43^\circ)=\frac{h}{x-18} \Rightarrow h=(x-18)\tan(43^\circ)$$

Step 3: Set equal and solve

$$x\tan(28^\circ)=(x-18)\tan(43^\circ)$$

Use decimal values:

$$x(0.5317)=(x-18)(0.9325)$$

$$0.5317x=0.9325x-16.785$$

$$-0.4008x=-16.785$$

$$x\approx 41.88$$

Step 4: Find the height

$$h=x\tan(28^\circ)$$

$$h\approx 41.88(0.5317)\approx 22.27$$

Answer: The tree is about \(22.3\) m tall.

8. When the observer’s eye level matters

Sometimes the angle of elevation is measured from a person’s eyes, not from the ground. If the person’s eye level is 1.6 m above the ground, then the vertical side in the triangle is not the full height of the object.

If a pole has total height \(H\), then the vertical side seen from eye level is:

$$H-1.6$$

This detail is important in more realistic problems.

Worked Example 4: Including eye level

From a point on level ground, a person with eye level 1.5 m measures the angle of elevation to the top of a flagpole as \(32^\circ\). After walking 12 m closer, the angle becomes \(47^\circ\). Find the total height of the flagpole.

Step 1: Define variables

• Let \(x\) be the original distance from the flagpole.
• The new distance is \(x-12\).
• Let \(H\) be the total height of the flagpole.

The vertical side of each triangle is:

$$H-1.5$$

Step 2: Write equations

$$\tan(32^\circ)=\frac{H-1.5}{x}$$

$$\tan(47^\circ)=\frac{H-1.5}{x-12}$$

So:

$$H-1.5=x\tan(32^\circ)$$

$$H-1.5=(x-12)\tan(47^\circ)$$

Step 3: Set equal

$$x\tan(32^\circ)=(x-12)\tan(47^\circ)$$

Use decimal values:

$$x(0.6249)=(x-12)(1.0724)$$

$$0.6249x=1.0724x-12.8688$$

$$-0.4475x=-12.8688$$

$$x\approx 28.76$$

Step 4: Find \(H\)

$$H-1.5=x\tan(32^\circ)$$

$$H-1.5\approx 28.76(0.6249)\approx 17.98$$

$$H\approx 19.48$$

Answer: The flagpole is about \(19.5\) m tall.

9. How to decide which trig ratio to use

In most multi-triangle height and distance problems on level ground, tangent is best because:

• the height is the opposite side, and
• the ground distance is the adjacent side.

So the basic model is:

$$\tan(\theta)=\frac{\text{height}}{\text{distance}}$$

You may use sine or cosine if a hypotenuse is given, but most of these problems are easiest with tangent.

10. Common mistakes to avoid

• Using the wrong distance expression: If the observer moves farther away, use \(x+\text{amount}\). If the observer moves closer, use \(x-\text{amount}\).
• Matching the wrong angle with the wrong triangle: The closer point should usually have the larger angle of elevation.
• Forgetting eye level: If the angle is measured from a person’s eyes, the triangle height is the object’s height minus eye level.
• Rounding too early: Keep calculator values for trig ratios until the end if possible.
• Not checking reasonableness: A very tall object should usually produce a larger angle if you are close to it.

11. Quick problem-solving checklist

1. What is the same in both triangles? Usually the height.
2. What changes? Usually the horizontal distance.
3. What variable should represent the unknown distance?
4. Is the second distance \(x+\text{amount}\) or \(x-\text{amount}\)?
5. Can I write two tangent equations?
6. After solving, does the larger angle match the shorter distance?

12. Summary

Multi-triangle height and distance problems use two connected right triangles to find unknown measurements such as height and distance. The most common tool is the tangent ratio:

$$\tan(\theta)=\frac{\text{height}}{\text{distance}}$$

To solve these problems, define variables carefully, write one trig equation for each triangle, and use the fact that both triangles describe the same object. Pay close attention to whether the observer moves closer or farther away, and include eye level when needed.

With practice, these problems become much easier because the structure repeats: draw, label, write two tangent equations, solve, and check.