Similarity, Proportions, and Right Triangle Trigonometry

Dilations and Scale Factors

In geometry, a dilation is a transformation that changes the size of a figure without changing its shape. This means the new figure is a larger or smaller version of the original figure, but the angles stay the same and the sides stay proportional.

Dilations are an important part of similarity. If one figure is a dilation of another, then the two figures are similar. They have the same shape, but not necessarily the same size.

A dilation uses two important ideas:

• a center of dilation
• a scale factor

The center of dilation is the fixed point from which the figure grows or shrinks. In many problems, the center of dilation is the origin, \, \((0,0)\).

The scale factor tells how much the figure is enlarged or reduced.

If the scale factor is:

• greater than 1, the figure gets larger. This is called an enlargement.
• between 0 and 1, the figure gets smaller. This is called a reduction.
• equal to 1, the figure stays the same size.

For example:

• A scale factor of \(2\) doubles every distance from the center.
• A scale factor of \(\frac{1}{2}\) cuts every distance in half.
• A scale factor of \(3\) triples every distance from the center.

How coordinates change in a dilation from the origin

When the center of dilation is the origin, each coordinate is multiplied by the scale factor.

If the original point is \((x,y)\) and the scale factor is \(k\), then the image is:

This rule makes dilations easier to work with on the coordinate plane.

What stays the same and what changes

In a dilation:

• the shape stays the same
• the angle measures stay the same
• the side lengths change by the scale factor
• the perimeter changes by the scale factor

So if a side length is multiplied by \(k\), then every other side length is also multiplied by \(k\).

For example, if a triangle has side lengths \(3\), \(4\), and \(5\), and it is dilated by a scale factor of \(2\), then the new side lengths are:

The triangle is bigger, but it still has the same shape.

Finding the scale factor

If you know a side length of the original figure and the matching side length of the image, you can find the scale factor using:

This ratio must be the same for all matching sides in similar figures.

Worked Example 1: Dilating a single point

Point \(A(3,4)\) is dilated from the origin by a scale factor of \(2\). Find the image of the point.

Use the rule:

Substitute \((3,4)\):

So the image is \(A'(6,8)\).

This makes sense because the point is now twice as far from the origin in both the \(x\)-direction and the \(y\)-direction.

Worked Example 2: Reducing a figure on the coordinate plane

Triangle \(ABC\) has vertices:

• \(A(2,6)\)
• \(B(4,2)\)
• \(C(6,6)\)

The triangle is dilated from the origin by a scale factor of \(\frac{1}{2}\). Find the image of each vertex.

Multiply each coordinate by \(\frac{1}{2}\):

So the image is:

• \(A'(1,3)\)
• \(B'(2,1)\)
• \(C'(3,3)\)

Because the scale factor is less than 1, the image is smaller than the original triangle.

Worked Example 3: Finding the scale factor from side lengths

A rectangle has a length of \(8\) cm and is dilated to a new length of \(12\) cm. What is the scale factor?

Use the formula:

Simplify:

So the scale factor is \(\frac{3}{2}\) or 1.5.

This means every side of the rectangle is multiplied by \(1.5\).

If the original width was \(4\) cm, the new width would be:

Worked Example 4: Using coordinates and side lengths together

Triangle \(PQR\) has vertices \(P(1,1)\), \(Q(3,1)\), and \(R(1,5)\). It is dilated from the origin by a scale factor of \(3\).

Step 1: Find the image of each vertex.

Step 2: Check one side length.

The original segment \(PQ\) goes from \((1,1)\) to \((3,1)\), so its length is:

The image segment \(P'Q'\) goes from \((3,3)\) to \((9,3)\), so its length is:

Compare the lengths:

This matches the scale factor. The side length was multiplied by \(3\), just as expected.

Important ideas to remember

• A dilation changes size, not shape.
• If the center is the origin, multiply both coordinates by the scale factor.
• A scale factor greater than 1 makes the figure larger.
• A scale factor between 0 and 1 makes the figure smaller.
• Matching side lengths in similar figures form equal ratios.

Common mistakes

• Adding instead of multiplying: In a dilation, you multiply coordinates by the scale factor. Do not add the scale factor.
• Using the wrong ratio: To find scale factor, divide image length by original length.
• Forgetting both coordinates: When dilating from the origin, multiply both \(x\) and \(y\).
• Thinking the shape changes: The size changes, but the figure stays similar to the original.

Quick practice questions

1. What is the image of \((5,2)\) after a dilation from the origin with scale factor \(2\)?
2. What is the image of \((8,4)\) after a dilation from the origin with scale factor \(\frac{1}{2}\)?
3. A side length changes from \(10\) to \(15\). What is the scale factor?
4. A triangle is dilated by a scale factor of \(4\). If one side was \(3\), what is the new side length?

Answers

1. \((10,4)\)
2. \((4,2)\)
3. \(\frac{15}{10}=\frac{3}{2}\)
4. \(3 \cdot 4 = 12\)

Summary

A dilation is a transformation that makes a figure larger or smaller while keeping the same shape. The amount of change is controlled by the scale factor, and when the center of dilation is the origin, you find new coordinates by multiplying each coordinate by the scale factor.

If the scale factor is greater than 1, the figure enlarges. If it is between 0 and 1, the figure reduces. In every dilation, matching side lengths are multiplied by the same number, which is why dilated figures are similar.

Similarity Postulates help us decide when two triangles have the same shape, even if they are different sizes.

In this lesson, you will learn how to prove triangles are similar using the three main similarity postulates: AA, SAS, and SSS. You will also learn how to check proportional side lengths and matching angles carefully.

This idea is important because similar triangles appear in geometry, scale drawings, maps, and right triangle trigonometry. When triangles are similar, we can use one triangle to find missing lengths in another.

What does similar mean?

Two figures are similar if they have the same shape but not necessarily the same size.

For triangles, this means:

• All corresponding angles are equal.
• All corresponding side lengths are proportional.

If triangle 1 and triangle 2 are similar, we write:

\(\triangle ABC \sim \triangle DEF\)

This statement tells us the matching order of the vertices:

• \(A \leftrightarrow D\)
• \(B \leftrightarrow E\)
• \(C \leftrightarrow F\)

So the corresponding sides are:

• \(AB \leftrightarrow DE\)
• \(BC \leftrightarrow EF\)
• \(AC \leftrightarrow DF\)

Similarity vs. Congruence

It is easy to confuse these ideas.

• Congruent figures have the same shape and the same size.
• Similar figures have the same shape, but sizes can be different.

So, every congruent triangle is also similar, but not every similar triangle is congruent.

The 3 Similarity Postulates

There are three main ways to prove that triangles are similar:

1. AA Similarity
2. SAS Similarity
3. SSS Similarity

Let’s study each one.

1. AA Similarity

AA stands for Angle-Angle.

If two angles of one triangle are equal to two angles of another triangle, then the triangles are similar.

This works because the third angles must also be equal. The angles in every triangle add up to \(180^\circ\).

So if:

\(\angle A = \angle D\) and \(\angle B = \angle E\)

then:

\(\triangle ABC \sim \triangle DEF\)

Why AA works:

If two triangles have the same angle measures, they must have the same shape. One triangle may just be a scaled-up or scaled-down version of the other.

2. SAS Similarity

SAS stands for Side-Angle-Side.

If:

• two pairs of corresponding sides are proportional, and
• the included angle between those sides is equal,

then the triangles are similar.

The included angle is the angle between the two sides you are comparing.

So if:

$$\frac{AB}{DE} = \frac{AC}{DF}$$

and

\(\angle A = \angle D\),

then:

\(\triangle ABC \sim \triangle DEF\)

Important: The equal angle must be between the two proportional sides. If the angle is not included, you cannot use SAS similarity.

3. SSS Similarity

SSS stands for Side-Side-Side.

If all three pairs of corresponding sides are proportional, then the triangles are similar.

So if:

$$\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$$

then:

\(\triangle ABC \sim \triangle DEF\)

Key idea: The side lengths do not need to be equal. They only need to have the same ratio.

How to check if sides are proportional

To test whether sides are proportional, write matching sides in the same order and compare their ratios.

For example, suppose one triangle has side lengths \(3, 4, 5\) and another has side lengths \(6, 8, 10\).

Check the ratios:

$$\frac{3}{6} = \frac{4}{8} = \frac{5}{10} = \frac{1}{2}$$

Since all three ratios are equal, the side lengths are proportional.

This means the triangles are similar by SSS.

Corresponding parts must match

One of the most common mistakes is matching the wrong sides or angles.

If:

\(\triangle ABC \sim \triangle DEF\)

then the order matters. This means:

• \(A\) matches \(D\)
• \(B\) matches \(E\)
• \(C\) matches \(F\)

So:

• \(AB\) matches \(DE\)
• \(BC\) matches \(EF\)
• \(AC\) matches \(DF\)

If you mix up the matching sides, your proportions will be wrong.

Worked Example 1: Using AA Similarity

Determine whether the triangles are similar.

Triangle 1 has angles \(50^\circ\), \(60^\circ\), and \(70^\circ\).

Triangle 2 has angles \(50^\circ\), \(70^\circ\), and \(60^\circ\).

Step 1: Compare angles.

Both triangles have the same three angle measures.

Step 2: Apply the postulate.

If two angles match, that is enough for AA similarity.

Conclusion:

The triangles are similar by AA Similarity.

Worked Example 2: Using SAS Similarity

Determine whether the triangles are similar.

In triangle 1, two sides are \(6\) and \(9\), and the included angle is \(40^\circ\).

In triangle 2, the corresponding sides are \(10\) and \(15\), and the included angle is also \(40^\circ\).

Step 1: Check the side ratios.

$$\frac{6}{10} = \frac{3}{5}$$

$$\frac{9}{15} = \frac{3}{5}$$

The side ratios are equal.

Step 2: Check the included angle.

Both included angles are \(40^\circ\).

Step 3: Apply SAS.

Two pairs of corresponding sides are proportional, and the included angle is equal.

Conclusion:

The triangles are similar by SAS Similarity.

Worked Example 3: Using SSS Similarity

Determine whether the triangles with side lengths \(4, 6, 8\) and \(6, 9, 12\) are similar.

Step 1: Compare the side ratios.

$$\frac{4}{6} = \frac{2}{3}$$

$$\frac{6}{9} = \frac{2}{3}$$

$$\frac{8}{12} = \frac{2}{3}$$

All three ratios are equal.

Step 2: Apply SSS.

Since all corresponding sides are proportional, the triangles are similar.

Conclusion:

The triangles are similar by SSS Similarity.

Worked Example 4: Decide which postulate works

Suppose in two triangles:

• \(\angle A = \angle D = 75^\circ\)
• \(AB = 8\), \(AC = 12\)
• \(DE = 10\), \(DF = 15\)

Are the triangles similar?

Step 1: Check the side ratios around the known angle.

$$\frac{AB}{DE} = \frac{8}{10} = \frac{4}{5}$$

$$\frac{AC}{DF} = \frac{12}{15} = \frac{4}{5}$$

The two side ratios are equal.

Step 2: Check the angle.

The angle between those sides is equal: \(\angle A = \angle D\).

Step 3: Choose the postulate.

This is SAS Similarity.

Conclusion:

The triangles are similar by SAS.

How similarity connects to scale factor

When triangles are similar, one triangle is a scaled version of the other.

The scale factor is the ratio of corresponding sides.

For example, if:

$$\frac{AB}{DE} = \frac{1}{2}$$

then triangle \(DEF\) is twice as large as triangle \(ABC\), or triangle \(ABC\) is half the size of triangle \(DEF\).

This idea helps when solving for missing side lengths.

Common mistakes to avoid

• Using the wrong order. Always match corresponding vertices and sides in the same order.
• Using non-matching sides in a proportion. Ratios must compare corresponding sides only.
• Forgetting the included angle in SAS. The equal angle must be between the two proportional sides.
• Thinking equal side lengths are required. For similarity, side lengths only need to be proportional, not equal.
• Checking only two sides for SSS. SSS needs all three pairs of sides proportional.

Quick guide: Which postulate should I use?

• If you know two angles, use AA.
• If you know two sides in proportion and the included angle, use SAS.
• If you know all three pairs of sides are proportional, use SSS.

Why this matters in right triangles

In right triangle trigonometry, similar triangles help explain why ratios like sine, cosine, and tangent stay the same for equal angles.

If two right triangles share the same acute angle, they are similar by AA. That means the ratios of their corresponding sides are equal.

This is why trig ratios work.

Summary

To prove triangles are similar, use one of the three similarity postulates:

• AA: two pairs of corresponding angles are equal.
• SAS: two pairs of corresponding sides are proportional and the included angle is equal.
• SSS: all three pairs of corresponding sides are proportional.

Always match corresponding parts carefully and check proportions in the correct order. Once triangles are proven similar, you can use their side ratios to solve for missing lengths and understand scale relationships.

Side-Splitter and Triangle Proportionality Theorems

When a line cuts across a triangle and is parallel to one side, it creates smaller parts that follow special ratio rules. These rules help us find missing side lengths without measuring every piece directly.

In this lesson, you will learn two closely related ideas:

• Triangle Proportionality Theorem
• Side-Splitter Theorem

These theorems are based on the idea of similar triangles. If two triangles have the same shape, then their matching sides are proportional.

1. Big Idea: A Parallel Line Creates Proportions

Imagine triangle \(ABC\). A line passes through sides \(AB\) and \(AC\), meeting them at points \(D\) and \(E\). If \(DE\) is parallel to \(BC\), then the smaller triangle \(ADE\) has the same shape as triangle \(ABC\).

Because \(DE \parallel BC\), corresponding angles are equal, so triangles \(ADE\) and \(ABC\) are similar.

That means:

This is the main reason the side-splitter relationships work.

2. Triangle Proportionality Theorem

The Triangle Proportionality Theorem says:

If a line is parallel to one side of a triangle and intersects the other two sides, then it divides those two sides proportionally.

Using triangle \(ABC\) with \(DE \parallel BC\):

This means the line splits side \(AB\) and side \(AC\) in the same ratio.

So if one side is split into a ratio of \(2:3\), the other side is also split into a ratio of \(2:3\).

3. Side-Splitter Theorem

The Side-Splitter Theorem is another name often used for this same proportional idea inside a triangle.

If a segment parallel to one side of a triangle cuts the other two sides, then the two sides are split proportionally:

Some teachers use the names Triangle Proportionality Theorem and Side-Splitter Theorem almost interchangeably. For 9th grade, the most important thing is to recognize the picture and know that parallel lines inside a triangle create equal ratios.

4. Important Notes Before Solving

• The inside segment must be parallel to one side of the triangle.
• Make sure you match the correct side parts.
• You can write proportions using part-to-part or small-to-whole ratios, but stay consistent.
• If two triangles are similar, corresponding sides must match in the correct order.

5. Common Ratio Forms

If \(DE \parallel BC\) in triangle \(ABC\), then these are all valid proportional relationships:

These all come from triangle similarity. Different problems may be easier with different forms.

6. Worked Example 1: Find a Missing Piece Using Part-to-Part Ratios

In triangle \(ABC\), points \(D\) and \(E\) lie on sides \(AB\) and \(AC\). Suppose \(DE \parallel BC\), \(AD=4\), \(DB=6\), and \(AE=8\). Find \(EC\).

Since \(DE\) is parallel to \(BC\), use the Triangle Proportionality Theorem:

Substitute the values:

Cross multiply:

Answer: \(EC=12\)

7. Worked Example 2: Use Small-to-Whole Ratios

In triangle \(ABC\), \(DE \parallel BC\). Let \(AD=5\), \(AB=15\), and \(AC=21\). Find \(AE\).

Because triangles \(ADE\) and \(ABC\) are similar, use:

Substitute:

Simplify \(\frac{5}{15}\) to \(\frac{1}{3}\):

Cross multiply:

Answer: \(AE=7\)

8. Worked Example 3: Solve for a Variable

In triangle \(PQR\), segment \(ST\) is parallel to \(QR\). Point \(S\) is on \(PQ\), and point \(T\) is on \(PR\). Suppose \(PS=x\), \(SQ=9\), \(PT=6\), and \(TR=18\). Find \(x\).

Use the side-splitter ratio:

Substitute the values:

Simplify \(\frac{6}{18}\) to \(\frac{1}{3}\):

Multiply both sides by 9:

Answer: \(x=3\)

9. Worked Example 4: Multi-Step Problem

In triangle \(LMN\), segment \(XY\) is parallel to \(MN\). Point \(X\) is on \(LM\), and point \(Y\) is on \(LN\). Suppose \(LX=8\), \(XM=4\), and the whole side \(LN=18\). Find \(LY\).

First, find the whole length of \(LM\):

Now use the similar triangle ratio:

Substitute:

Simplify \(\frac{8}{12}\) to \(\frac{2}{3}\):

Cross multiply or think of \(\frac{2}{3}\) of 18:

Answer: \(LY=12\)

10. How to Recognize These Problems

You are probably working with side-splitter or triangle proportionality if:

• You see a triangle with a segment drawn inside it.
• The inside segment is marked parallel to one side.
• The problem asks for a missing side length.
• The side lengths are split into smaller pieces.

11. Step-by-Step Strategy

1. Check that the segment inside the triangle is parallel to a side.
2. Identify the smaller triangle and the larger triangle.
3. Choose a proportion that matches the information given.
4. Substitute carefully.
5. Solve using cross multiplication.
6. Check whether your answer makes sense in the diagram.

12. Common Mistakes to Avoid

• Using the theorem without parallel lines: The segment must be parallel to one side of the triangle.
• Mixing up matching sides: Keep corresponding sides in the same order.
• Using part with whole incorrectly: If one fraction is part-to-whole, the other must also be part-to-whole.
• Forgetting to add pieces: Sometimes you need the whole side, so add the side parts first.

13. Quick Check

If \(DE \parallel BC\), \(AD=3\), \(DB=5\), and \(AE=6\), what is \(EC\)?

Use:

14. Summary

The Triangle Proportionality Theorem and Side-Splitter Theorem tell us that when a line is parallel to one side of a triangle, it divides the other two sides proportionally.

This works because the small triangle and the large triangle are similar. Once you identify the matching sides, you can write a proportion and solve for any missing length.

Always check for parallel lines, match corresponding sides carefully, and decide whether you should use side parts or whole side lengths.

Pythagorean Theorem and Converse

When working with triangles, one very important special case is the right triangle. A right triangle has one angle that measures exactly 900degrees.

The Pythagorean Theorem helps us find missing side lengths in right triangles. Its reverse idea, called the converse, helps us decide whether a triangle is right, acute, or obtuse by looking only at the side lengths.

This lesson will show you how to use both ideas step by step.

1. Parts of a Right Triangle

In a right triangle, the two sides that form the right angle are called the legs. The side across from the right angle is called the hypotenuse.

• The hypotenuse is always the longest side.
• The other two sides are the legs.

If the legs are named \(a\) and \(b\), and the hypotenuse is named \(c\), then the theorem is:

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

This means that the sum of the squares of the legs equals the square of the hypotenuse.

2. When to Use the Pythagorean Theorem

Use the Pythagorean Theorem when:

• the triangle is a right triangle, and
• you know two side lengths and need to find the third.

You can use it to find:

• the hypotenuse if both legs are known, or
• a missing leg if one leg and the hypotenuse are known.

3. Finding the Hypotenuse

If both legs are known, substitute them into the formula and solve for \(c\).

Worked Example 1: Find the hypotenuse of a right triangle with legs 6 and 8.

Start with the formula:

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

Substitute \(a=6\) and \(b=8\):

$$6^2+8^2=c^2$$

Square each number:

$$36+64=c^2$$

Add:

$$100=c^2$$

Take the square root of both sides:

$$c=10$$

So the hypotenuse is 10.

4. Finding a Missing Leg

If you know the hypotenuse and one leg, you can find the other leg by subtracting first.

Worked Example 2: A right triangle has hypotenuse 13 and one leg 5. Find the other leg.

Use the formula:

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

Let the missing leg be \(b\). Substitute:

$$5^2+b^2=13^2$$

Square the known numbers:

$$25+b^2=169$$

Subtract 25 from both sides:

$$b^2=144$$

Take the square root:

$$b=12$$

So the missing leg is 12.

5. Exact Answers and Simplified Radicals

Sometimes the square root does not come out to a whole number. In that case, leave the answer in simplified radical form if asked for an exact answer.

Worked Example 3: Find the hypotenuse of a right triangle with legs 5 and 7.

Use the theorem:

$$5^2+7^2=c^2$$

$$25+49=c^2$$

$$74=c^2$$

$$c=\sqrt{74}$$

Since 74 does not have a perfect-square factor, this is already simplified.

So the exact hypotenuse is \(\sqrt{74}\).

If a decimal approximation is needed, then:

$$\sqrt{74}\approx 8.6$$

6. Common Pythagorean Triples

Some right triangles have side lengths that are all whole numbers. These are called Pythagorean triples.

• \(3,4,5\)
• \(5,12,13\)
• \(8,15,17\)

Multiples of these also work:

• \(6,8,10\) is a multiple of \(3,4,5\)
• \(10,24,26\) is a multiple of \(5,12,13\)

Recognizing these can help you solve problems faster.

7. The Converse of the Pythagorean Theorem

The converse works in reverse. Instead of starting with a right triangle, you start with three side lengths and test what kind of triangle they make.

First, identify the longest side. Call it \(c\). Then compare \(a^2+b^2\) with \(c^2\).

• If \(a^2+b^2=c^2\), the triangle is right.
• If \(a^2+b^2>c^2\), the triangle is acute.
• If \(a^2+b^2<c^2\), the triangle is obtuse.

This lets you classify a triangle using side lengths only.

8. Using the Converse to Classify Triangles

Worked Example 4: Classify the triangle with side lengths 7, 24, and 25.

The longest side is 25, so let \(c=25\).

Now compare:

$$7^2+24^2 \quad \text{and} \quad 25^2$$

Compute each side:

$$49+576=625$$

$$25^2=625$$

Since:

$$7^2+24^2=25^2$$

the triangle is a right triangle.

One more quick classification example: Classify the triangle with side lengths 6, 8, and 11.

The longest side is 11.

$$6^2+8^2=36+64=100$$

$$11^2=121$$

Since:

$$100<121$$

we have \(a^2+b^2<c^2\), so the triangle is obtuse.

9. Important Tips

• Only use the Pythagorean Theorem to find a missing side if the triangle is right.
• The hypotenuse must always be the longest side.
• When solving for a missing leg, subtract before taking the square root.
• When using the converse, always compare the two shorter sides with the longest side.
• Check your arithmetic carefully when squaring numbers.

10. Summary

The Pythagorean Theorem is:

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

It is used to find missing side lengths in a right triangle.

The converse of the Pythagorean Theorem is used to classify triangles from their side lengths:

• \(a^2+b^2=c^2\) means right
• \(a^2+b^2>c^2\) means acute
• \(a^2+b^2<c^2\) means obtuse

If you remember to identify the longest side first and use the formula carefully, you can solve many triangle problems with confidence.

Special Right Triangles are right triangles with side lengths that follow fixed patterns. These patterns let you find missing sides quickly without using a calculator.

In 9th Grade Maths, the two most important special right triangles are:

• 45-45-90 triangles
• 30-60-90 triangles

These triangles appear often in geometry, similarity, and trigonometry. If you recognize them, you can solve problems much faster.

Before we begin: a right triangle has one angle that measures \(90^\circ\). The side across from the right angle is called the hypotenuse. The other two sides are called legs.

1. The 45-45-90 Triangle

A 45-45-90 triangle is a right triangle with the other two angles both equal to \(45^\circ\). Since the two acute angles are equal, the triangle is also an isosceles triangle, so its two legs are equal in length.

The side ratio in a 45-45-90 triangle is:

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

This means:

• the two legs are equal
• the hypotenuse is the leg multiplied by \(\sqrt{2}\)

If each leg has length \(x\), then the sides are:

$$x,\ x,\ x\sqrt{2}$$

So the main formulas are:

$$\text{hypotenuse} = (\text{leg})\sqrt{2}$$

$$\text{leg} = \frac{\text{hypotenuse}}{\sqrt{2}}$$

You may also rationalize the denominator:

$$\frac{x}{\sqrt{2}} = \frac{x\sqrt{2}}{2}$$

2. The 30-60-90 Triangle

A 30-60-90 triangle is a right triangle with angles \(30^\circ\), \(60^\circ\), and \(90^\circ\).

The side ratio in a 30-60-90 triangle is:

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

These side lengths match the angles across from them:

• the side opposite \(30^\circ\) is the short leg
• the side opposite \(60^\circ\) is the long leg
• the side opposite \(90^\circ\) is the hypotenuse

If the short leg is \(x\), then the sides are:

$$x,\ x\sqrt{3},\ 2x$$

So the main formulas are:

$$\text{long leg} = (\text{short leg})\sqrt{3}$$

$$\text{hypotenuse} = 2(\text{short leg})$$

$$\text{short leg} = \frac{\text{hypotenuse}}{2}$$

Important: In a 30-60-90 triangle, the shortest side is always opposite the \(30^\circ\) angle.

3. How to Recognize a Special Right Triangle

You may know a triangle is special in different ways:

• its angles are given, such as \(45^\circ,45^\circ,90^\circ\) or \(30^\circ,60^\circ,90^\circ\)
• its side lengths already match a known ratio
• it is formed from a familiar shape, such as a square or an equilateral triangle

For example:

• cutting a square along a diagonal makes two 45-45-90 triangles
• cutting an equilateral triangle in half makes two 30-60-90 triangles

4. Why These Ratios Work

You do not need to memorize where they come from in great detail, but it helps to know the idea.

For a 45-45-90 triangle, start with a square of side length \(x\). The diagonal is the hypotenuse. Using the Pythagorean Theorem:

$$x^2+x^2=c^2$$

$$2x^2=c^2$$

$$c=x\sqrt{2}$$

So the ratio is \(1:1:\sqrt{2}\).

For a 30-60-90 triangle, start with an equilateral triangle of side length \(2x\). Split it in half. The half-triangle has:

• short leg \(x\)
• hypotenuse \(2x\)
• long leg found by the Pythagorean Theorem

$$x^2+(\text{long leg})^2=(2x)^2$$

$$x^2+(\text{long leg})^2=4x^2$$

$$(\text{long leg})^2=3x^2$$

$$\text{long leg}=x\sqrt{3}$$

So the ratio is \(1:\sqrt{3}:2\).

5. Worked Examples

Example 1: Find the hypotenuse in a 45-45-90 triangle

A 45-45-90 triangle has legs of length \(8\). Find the hypotenuse.

Step 1: Use the 45-45-90 rule.

$$\text{hypotenuse} = (\text{leg})\sqrt{2}$$

Step 2: Substitute \(8\) for the leg.

$$\text{hypotenuse} = 8\sqrt{2}$$

Answer: The hypotenuse is \(8\sqrt{2}\).

Example 2: Find a leg in a 45-45-90 triangle

A 45-45-90 triangle has hypotenuse \(12\). Find the length of each leg.

Step 1: Use the formula for a leg.

$$\text{leg} = \frac{\text{hypotenuse}}{\sqrt{2}}$$

Step 2: Substitute \(12\).

$$\text{leg} = \frac{12}{\sqrt{2}}$$

Step 3: Rationalize the denominator.

$$\text{leg} = \frac{12\sqrt{2}}{2}=6\sqrt{2}$$

Answer: Each leg is \(6\sqrt{2}\).

Example 3: Find the missing sides in a 30-60-90 triangle

In a 30-60-90 triangle, the short leg is \(5\). Find the long leg and the hypotenuse.

Step 1: Write the side pattern.

$$x,\ x\sqrt{3},\ 2x$$

Step 2: Let \(x=5\).

• short leg = \(5\)
• long leg = \(5\sqrt{3}\)
• hypotenuse = \(10\)

Answer: The long leg is \(5\sqrt{3}\), and the hypotenuse is \(10\).

Example 4: Find the missing side when the long leg is given

A 30-60-90 triangle has a long leg of \(9\sqrt{3}\). Find the short leg and the hypotenuse.

Step 1: In a 30-60-90 triangle,

$$\text{long leg}=(\text{short leg})\sqrt{3}$$

Step 2: Set up the equation.

$$9\sqrt{3}=x\sqrt{3}$$

Step 3: Divide both sides by \(\sqrt{3}\).

$$x=9$$

So the short leg is \(9\).

Step 4: Find the hypotenuse.

$$\text{hypotenuse}=2x=18$$

Answer: The short leg is \(9\), and the hypotenuse is \(18\).

6. Common Mistakes to Avoid

• Mixing up the triangle types: \(1:1:\sqrt{2}\) is only for 45-45-90 triangles, and \(1:\sqrt{3}:2\) is only for 30-60-90 triangles.
• Confusing the short leg and long leg: in a 30-60-90 triangle, the short leg is opposite \(30^\circ\).
• Adding instead of multiplying: the hypotenuse in a 45-45-90 triangle is not leg + \(\sqrt{2}\); it is leg \(\times \sqrt{2}\).
• Forgetting to simplify radicals: leave answers like \(8\sqrt{2}\) or \(5\sqrt{3}\) in simplest radical form.

7. Quick Reference

45-45-90 triangle

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

• legs are equal
• hypotenuse = leg \(\times \sqrt{2}\)

30-60-90 triangle

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

• short leg is opposite \(30^\circ\)
• long leg = short leg \(\times \sqrt{3}\)
• hypotenuse = short leg \(\times 2\)

8. When to Use Special Right Triangles

Use these patterns when:

• the triangle has angles \(45^\circ,45^\circ,90^\circ\)
• the triangle has angles \(30^\circ,60^\circ,90^\circ\)
• you want an exact answer with radicals instead of a decimal approximation

These triangles are especially useful in geometry problems, proofs, and trigonometry.

Summary

Special right triangles follow fixed side ratios that make solving for missing sides much easier. A 45-45-90 triangle has side ratio \(1:1:\sqrt{2}\), and a 30-60-90 triangle has side ratio \(1:\sqrt{3}:2\).

If you can identify which special triangle you have, then you can use the correct ratio to find missing sides quickly and exactly. Learning these two patterns will help you in many geometry and trigonometry problems.

Lesson: Sine, Cosine, and Tangent Ratios

When we work with right triangles, we can use special ratios to connect an angle to the side lengths of the triangle. These ratios are called sine, cosine, and tangent.

These ratios help us answer questions like:

• How tall is a tree if I know how far away I am and the angle I look up?
• How long is a ramp if I know its height and steepness?
• What side length is missing in a right triangle?

This lesson will teach you what sine, cosine, and tangent mean, how to remember them using SOH CAH TOA, and how to use them to solve problems.

1. Start with a right triangle

A right triangle has one angle that measures exactly \(90^\circ\). The other two angles are acute angles, which means they are less than \(90^\circ\).

To use trigonometric ratios, we pick one of the acute angles and compare the sides based on that angle.

The three sides of a right triangle are named like this:

• Hypotenuse: the longest side, opposite the right angle
• Opposite side: the side directly across from the chosen angle
• Adjacent side: the side next to the chosen angle that is not the hypotenuse

Important: The names opposite and adjacent depend on which acute angle you are looking at. The hypotenuse never changes.

2. The three trigonometric ratios

The three basic ratios are:

Here, \(\theta\) is the angle you are using.

A common memory trick is SOH CAH TOA:

• SOH: Sine = Opposite / Hypotenuse
• CAH: Cosine = Adjacent / Hypotenuse
• TOA: Tangent = Opposite / Adjacent

3. How to identify the sides

Suppose you choose an acute angle in a right triangle.

1. Find the hypotenuse first. It is always across from the right angle.
2. Look at the chosen angle.
3. The side across from that angle is the opposite side.
4. The side next to that angle, which is not the hypotenuse, is the adjacent side.

If you mix up the sides, your trigonometric ratio will be wrong. So always label the triangle carefully before doing any calculations.

4. Why these ratios work

Right triangles with the same angle measures are similar. That means their matching sides are proportional.

So, if two right triangles both have an angle of \(35^\circ\), the ratio of opposite to hypotenuse will always be the same. That constant ratio is called the sine of \(35^\circ\).

In the same way, the ratio of adjacent to hypotenuse is the cosine, and the ratio of opposite to adjacent is the tangent.

5. Worked Example 1: Find a trig ratio from side lengths

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

• opposite = 3
• adjacent = 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: Choose the correct ratio to find a missing side

A right triangle has an angle of \(40^\circ\). The hypotenuse is 10 cm. Find the side opposite the \(40^\circ\) angle.

Step 1: Decide which ratio uses opposite and hypotenuse.

That is sine.

Step 2: Substitute what you know.

Step 3: Solve for \(x\).

Using a calculator:

Answer: The opposite side is about \(6.43\) cm.

7. Worked Example 3: Use tangent to find a missing side

A ladder leans against a wall. The angle between the ladder and the ground is \(65^\circ\). The bottom of the ladder is 4 m from the wall. How high up the wall does the ladder reach?

Step 1: Identify the sides compared to the \(65^\circ\) angle.

• The height up the wall is the opposite side.
• The 4 m on the ground is the adjacent side.

Step 2: Choose the ratio with opposite and adjacent.

That is tangent.

Step 3: Solve.

Using a calculator:

Answer: The ladder reaches about \(8.58\) m up the wall.

8. Worked Example 4: Find an angle using a trig ratio

In a right triangle, the opposite side is 7 and the adjacent side is 10. Find the angle \(\theta\).

Step 1: Choose the ratio that uses opposite and adjacent.

Step 2: Use the inverse tangent on a calculator.

Answer: \(\theta \approx 35.0^\circ\)

9. How to know which ratio to use

Look at the sides you know and the side you need.

• If you use opposite and hypotenuse, use sine.
• If you use adjacent and hypotenuse, use cosine.
• If you use opposite and adjacent, use tangent.

A helpful strategy is:

1. Mark the given angle.
2. Label opposite, adjacent, and hypotenuse.
3. Circle the known side and the unknown side.
4. Pick the trig ratio that connects those two sides.

10. Calculator tips

When using sine, cosine, and tangent on a calculator, make sure it is in degree mode, not radian mode, because your angles in 9th Grade are usually given in degrees.

Examples:

• To find \(\sin(40^\circ)\), type sin(40).
• To find an angle from a sine ratio, use the inverse button, often written as sin^-1 or asin.

11. Common mistakes to avoid

• Mixing up opposite and adjacent: These depend on the chosen angle.
• Forgetting which side is the hypotenuse: It is always opposite the right angle.
• Using the wrong ratio: Match the ratio to the sides involved.
• Calculator in the wrong mode: Use degrees.
• Rounding too early: Keep more digits until the final step.

12. Quick practice ideas

For each problem, ask yourself:

• What is the chosen angle?
• Which side is opposite?
• Which side is adjacent?
• Which side is the hypotenuse?
• Do I need sine, cosine, or tangent?

13. Summary

Sine, cosine, and tangent are ratios used with right triangles. They compare side lengths to an angle.

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

Remember SOH CAH TOA to keep the ratios straight. First label the sides carefully, then choose the correct ratio, and finally solve for the missing side or angle.

Inverse Trigonometric Functions help us find a missing angle in a right triangle when we know a ratio of side lengths.

In earlier trigonometry, you may have used sine, cosine, and tangent to find a missing side when an angle was known. Now we do the reverse: we know a side ratio and want the angle. That is why these are called inverse trigonometric functions.

The three inverse trigonometric functions you will use most often are:

• arcsine, written as \(\sin^{-1}\) or \(\arcsin\)
• arccosine, written as \(\cos^{-1}\) or \(\arccos\)
• arctangent, written as \(\tan^{-1}\) or \(\arctan\)

These functions answer questions like:

• “What angle has sine \(\frac{3}{5}\)?”
• “What angle has cosine \(0.8\)?”
• “What angle has tangent \(\frac{7}{4}\)?”

For example, if

then

This means: find the angle whose sine is \(\frac{3}{5}\).

Important: in this lesson, we are working with right triangles, so the angle we find will be an acute angle, meaning it is between \(0^\circ\) and \(90^\circ\).

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

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

So the inverse versions are used like this:

• $$\theta = \sin^{-1}\left(\frac{\text{opposite}}{\text{hypotenuse}}\right)$$
• $$\theta = \cos^{-1}\left(\frac{\text{adjacent}}{\text{hypotenuse}}\right)$$
• $$\theta = \tan^{-1}\left(\frac{\text{opposite}}{\text{adjacent}}\right)$$

How to choose which inverse trig function to use

Look at the two sides you know compared to the angle you are trying to find.

• If you know opposite and hypotenuse, use sine.
• If you know adjacent and hypotenuse, use cosine.
• If you know opposite and adjacent, use tangent.

You may remember this with the pattern:

• SOH: sine = opposite/hypotenuse
• CAH: cosine = adjacent/hypotenuse
• TOA: tangent = opposite/adjacent

Then use the inverse function to solve for the angle.

Using a calculator

To find an angle, use the \(\sin^{-1}\), \(\cos^{-1}\), or \(\tan^{-1}\) button on your calculator. On some calculators, these are written as \(\arcsin\), \(\arccos\), and \(\arctan\).

Make sure your calculator is in degree mode, not radian mode, unless your teacher says otherwise. In 9th Grade right triangle problems, answers are usually given in degrees.

Worked Example 1: Using arcsine

In a right triangle, an angle \(\theta\) has opposite side \(6\) and hypotenuse \(10\). Find \(\theta\).

Step 1: Choose the trig ratio.

We know opposite and hypotenuse, so use sine.

Step 2: Use the inverse function.

Step 3: Calculate.

Answer: \(\theta \approx 36.9^\circ\)

Worked Example 2: Using arccosine

In a right triangle, an angle \(x\) has adjacent side \(8\) and hypotenuse \(17\). Find \(x\).

Step 1: Choose the trig ratio.

We know adjacent and hypotenuse, so use cosine.

Step 2: Use the inverse function.

Step 3: Calculate.

Answer: \(x \approx 61.9^\circ\)

Worked Example 3: Using arctangent

A ladder leans against a wall. The bottom of the ladder is \(4\) meters from the wall, and the top touches the wall \(9\) meters above the ground. Find the angle \(\theta\) that the ladder makes with the ground.

Step 1: Identify the sides compared to the angle.

The angle is at the ground. The side opposite the angle is the height on the wall, \(9\). The side adjacent to the angle is the ground distance, \(4\).

Step 2: Use tangent.

Step 3: Use the inverse function.

Step 4: Calculate.

Answer: The ladder makes an angle of about \(66.0^\circ\) with the ground.

Worked Example 4: Finding another angle

In a right triangle, the hypotenuse is \(13\) and one leg is \(5\). Find the acute angle \(\theta\) opposite the side of length \(5\).

Step 1: Identify the known ratio.

The side opposite \(\theta\) is \(5\), and the hypotenuse is \(13\), so use sine.

Step 2: Apply inverse sine.

Step 3: Calculate.

Answer: \(\theta \approx 22.6^\circ\)

What if you need the other acute angle?

In a right triangle, the two acute angles always add to \(90^\circ\).

So if one acute angle is \(22.6^\circ\), the other is

This is helpful when a problem asks for a different angle than the one you first found.

Common mistakes to avoid

• Using the wrong sides. Always name sides relative to the angle you are finding.
• Choosing the wrong trig function. Check whether your known sides match sine, cosine, or tangent.
• Forgetting the inverse button. To find an angle, use \(\sin^{-1}\), \(\cos^{-1}\), or \(\tan^{-1}\), not just \(\sin\), \(\cos\), or \(\tan\).
• Calculator in the wrong mode. If your answer looks strange, check that the calculator is in degree mode.
• Rounding too early. Keep more digits until the final step, then round your final answer.

Quick problem-solving steps

1. Draw or study the right triangle carefully.
2. Mark the angle you need to find.
3. Label the known sides as opposite, adjacent, or hypotenuse.
4. Choose sine, cosine, or tangent.
5. Write the trig equation.
6. Use the inverse trig function to find the angle.
7. Round the answer if needed, usually to the nearest tenth of a degree.

Practice Check

Try these on your own:

• If \(\sin \theta = \frac{7}{10}\), find \(\theta\).
• If \(\cos x = \frac{12}{13}\), find \(x\).
• If \(\tan y = \frac{3}{2}\), find \(y\).

The setup should look like this:

• $$\theta = \sin^{-1}\left(\frac{7}{10}\right)$$
• $$x = \cos^{-1}\left(\frac{12}{13}\right)$$
• $$y = \tan^{-1}\left(\frac{3}{2}\right)$$

Summary

Inverse trigonometric functions are used to find missing angles in right triangles from known side ratios.

Use:

• \(\sin^{-1}\) for opposite and hypotenuse
• \(\cos^{-1}\) for adjacent and hypotenuse
• \(\tan^{-1}\) for opposite and adjacent

Always identify the sides correctly, choose the matching trig ratio, and make sure your calculator is in degree mode. With practice, inverse trig becomes a quick and powerful way to find unknown angles.

Angles of Elevation and Depression help us describe how we look upward or downward at an object. These ideas are used in real-life situations such as finding the height of a building, the distance to a boat from a cliff, or how high a drone is flying.

These problems are usually solved by drawing a right triangle and using trigonometric ratios such as sine, cosine, and tangent. In 9th Grade, the most common ratio for these questions is tangent, because it connects a height and a horizontal distance.

Angle of elevation is the angle formed when you look up from a horizontal line.

Angle of depression is the angle formed when you look down from a horizontal line.

In both cases, the angle is measured from a horizontal line, not from a vertical line.

Here is the key idea:

• If you are looking up, it is an angle of elevation.
• If you are looking down, it is an angle of depression.
• These situations often create a right triangle.
• Once the triangle is drawn, use the side lengths and the correct trigonometric ratio.

Important picture idea: A horizontal line is always straight across. The angle of elevation or depression is measured between that horizontal line and the line of sight.

When solving these problems, the triangle usually includes:

• a horizontal distance
• a vertical height
• a line of sight, which is often the slanted side

Because the vertical and horizontal lines are perpendicular, they form a right angle. That is why right triangle trigonometry works.

Useful trigonometric ratios:

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

For many angle of elevation and depression problems:

• opposite is the vertical height
• adjacent is the horizontal distance

So a very common equation is:

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

How to solve these problems

1. Draw the situation as a right triangle.
2. Mark the given angle carefully.
3. Label the known side and the unknown side.
4. Choose the trigonometric ratio that matches the sides you are using.
5. Substitute the values into the equation.
6. Solve for the unknown.
7. Check if your answer makes sense.

Very important fact: In angle of depression problems, the angle of depression is often equal to the angle of elevation in the triangle below. This happens because horizontal lines are parallel.

So even if the angle is given at the top of a building or cliff, you can often use the same angle inside the triangle at the ground level.

Worked Example 1: Finding a height from an angle of elevation

A student stands 20 m from a tree. The angle of elevation from the student to the top of the tree is \(35^\circ\). Find the height of the tree.

Step 1: Identify the sides.

• Horizontal distance = 20 m
• Height of tree = unknown
• Angle of elevation = \(35^\circ\)

Step 2: Choose a ratio.

The height is opposite the angle, and the 20 m is adjacent to the angle. So use tangent:

$$\tan(35^\circ)=\frac{h}{20}$$

Step 3: Solve.

$$h=20\tan(35^\circ)$$

$$h\approx 20(0.7002)$$

$$h\approx 14.0$$

Answer: The tree is about 14.0 m tall.

Worked Example 2: Finding a horizontal distance

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 \(50^\circ\). Find the horizontal distance from the person to the point directly below the kite.

Step 1: Identify the sides.

• Opposite side = 18 m
• Adjacent side = \(d\)
• Angle = \(50^\circ\)

Step 2: Use tangent.

$$\tan(50^\circ)=\frac{18}{d}$$

Step 3: Solve for \(d\).

Multiply both sides by \(d\):

$$d\tan(50^\circ)=18$$

Now divide by \(\tan(50^\circ)\):

$$d=\frac{18}{\tan(50^\circ)}$$

$$d\approx \frac{18}{1.1918}$$

$$d\approx 15.1$$

Answer: The horizontal distance is about 15.1 m.

Worked Example 3: Angle of depression

A lighthouse is 40 m tall. The angle of depression from the top of the lighthouse to a boat is \(28^\circ\). How far is the boat from the base of the lighthouse?

Step 1: Understand the angle.

The angle of depression is measured downward from the horizontal at the top of the lighthouse. That angle is equal to the angle of elevation from the boat to the top of the lighthouse, so we can use \(28^\circ\) in the triangle.

Step 2: Label the triangle.

• Height = 40 m
• Horizontal distance = \(x\)
• Angle = \(28^\circ\)

Step 3: Use tangent.

$$\tan(28^\circ)=\frac{40}{x}$$

Step 4: Solve.

$$x=\frac{40}{\tan(28^\circ)}$$

$$x\approx \frac{40}{0.5317}$$

$$x\approx 75.2$$

Answer: The boat is about 75.2 m from the base of the lighthouse.

Worked Example 4: Using the hypotenuse

A drone is seen at an angle of elevation of \(62^\circ\). The line of sight from the observer to the drone is 30 m. Find the drone’s height above the ground.

Step 1: Identify the sides.

• Angle = \(62^\circ\)
• Hypotenuse = 30 m
• Opposite side = height \(h\)

Step 2: Choose the ratio.

We know the opposite side and the hypotenuse, so use sine:

$$\sin(62^\circ)=\frac{h}{30}$$

Step 3: Solve.

$$h=30\sin(62^\circ)$$

$$h\approx 30(0.8829)$$

$$h\approx 26.5$$

Answer: The drone is about 26.5 m above the ground.

Common mistakes to avoid

• Measuring from the wrong line: The angle of elevation or depression is measured from a horizontal line.
• Using the wrong trigonometric ratio: Check which sides you know: opposite, adjacent, or hypotenuse.
• Mixing up height and distance: Vertical height and horizontal distance are different sides.
• Forgetting the equal angles idea: In angle of depression problems, the angle of depression often equals the angle of elevation.
• Calculator mode errors: Make sure your calculator is in degree mode, not radian mode.

Quick check: Which ratio should I use?

• If you need height and horizontal distance, use tangent:

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

• If you need height and line of sight, use sine:

$$\sin(\theta)=\frac{\text{height}}{\text{line of sight}}$$

• If you need distance and line of sight, use cosine:

$$\cos(\theta)=\frac{\text{distance}}{\text{line of sight}}$$

Summary

Angles of elevation and depression describe how we look up or down from a horizontal line. These situations can be modeled with right triangles. After drawing the triangle, label the sides carefully and use sine, cosine, or tangent to find missing heights, distances, or line-of-sight lengths.

If you remember to start with a diagram, identify the angle correctly, and choose the matching trigonometric ratio, these problems become much easier to solve.