T5 Find right-triangle sides and angles using ratios and Pythagoras

Ontario Grade 10 Mathematics

Trigonometry

Use Pythagoras and trigonometric ratios to solve missing parts

A right triangle has one angle that measures

90^\circ

. If you know some of its side lengths or angles, you can use relationships between the sides to find what is missing. This lesson reviews the Pythagorean theorem and introduces three side-to-side ratios: sine, cosine, and tangent. A calculator can help with the ratios, but you still need to choose the right relationship and label the triangle carefully.

What you will learn

Identify the hypotenuse and the opposite and adjacent sides for a chosen angle.
Use the Pythagorean theorem to find a missing side in a right triangle.
Choose sine, cosine, or tangent to find a missing side or angle.
Round answers appropriately and check that they make sense.

1. Start with the triangle and its side names

A useful Grade 9 review is that a triangle’s three interior angles add to

180^\circ

. In a right triangle, one angle is

90^\circ

, so the other two angles must add to

90^\circ

. The two shorter sides meet at the right angle. The longest side is across from it.

The longest side is called the hypotenuse. It has the same name no matter which of the two smaller angles you are using. For a chosen acute angle, the opposite side is directly across from that angle. The adjacent side touches that angle and is not the hypotenuse. Opposite and adjacent can change when you choose the other acute angle.

Before using a ratio, mark the angle you are working with. Then label the hypotenuse, opposite, and adjacent sides from that angle. This prevents a common mix-up: calling a side opposite when it is actually adjacent to the marked angle.

A+B=90^\circ

The hypotenuse is always opposite the $90^\circ$ angle.
Opposite and adjacent are named in relation to the selected acute angle.
The two acute angles in a right triangle add to $90^\circ$ .

2. Find a missing side with Pythagoras

The Pythagorean theorem connects the three side lengths of a right triangle. If the legs are

a

and

b

, and the hypotenuse is

c

, then the squares of the leg lengths add to the square of the hypotenuse. A square means a number multiplied by itself.

For example, if the legs are

3

and

4

, then the hypotenuse is

5

, because

3^2+4^2=5^2

. This is a way to find a missing side when you know the other two sides. If the missing side is a leg, rearrange the equation by subtracting the known leg’s square before taking the square root.

Use the theorem only with a right triangle. Check first that the side opposite the right angle is the hypotenuse. If the hypotenuse is missing, the answer should be longer than either leg. If a leg is missing, it should be shorter than the hypotenuse.

a^2+b^2=c^2

Use the Pythagorean theorem for a right triangle.
The hypotenuse is the side represented by $c$ in the formula.
Take the positive square root because a side length cannot be negative.

3. Use sine, cosine, and tangent

A ratio compares two quantities by division. In a right triangle, each acute angle has three useful side ratios. Sine compares opposite with hypotenuse. Cosine compares adjacent with hypotenuse. Tangent compares opposite with adjacent. The short names are sin, cos, and tan.

Choose a ratio that contains the side you know and the side or angle you need. For instance, if you know an angle and the hypotenuse and need the opposite side, use sine. If you know an angle and the adjacent side and need the opposite side, use tangent. Keep the selected angle in mind when labelling sides.

To find an angle from two side lengths, use an inverse ratio. On many calculators, the inverse sine, inverse cosine, and inverse tangent functions are labelled

\sin^{-1}

,

\cos^{-1}

, and

\tan^{-1}

. These functions return an angle; they do not mean that you should divide by sine, cosine, or tangent. Set the calculator to degree mode when working with angles in degrees.

A sketch or a small side-name table can help you choose the ratio. Once you write the equation, substitute the known values and solve for the unknown. Round only at the end when possible, so early rounding does not affect later calculations.

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

Sine uses opposite and hypotenuse.
Cosine uses adjacent and hypotenuse.
Tangent uses opposite and adjacent.
Inverse ratios can find an angle from two known sides.

4. Guided example and independent practice

When solving a triangle, decide what information is already known and what is missing. If side lengths are involved, Pythagoras can find a third side. If an angle and side are involved, use a trigonometric ratio. If you know two side lengths and need an angle, use an inverse ratio. Your answer should fit the triangle and the information given.

Try these independently after studying the worked example. For each problem, sketch and label a right triangle before calculating. (1) A right triangle has legs of

7

cm and

24

cm. Find its hypotenuse. (2) For a marked angle, the opposite side is

6

cm and the hypotenuse is

10

cm. Find the angle to the nearest degree. (3) The hypotenuse is

15

cm and one acute angle is

30^\circ

. Find the opposite side to the nearest tenth.

To check your approach, ask whether your chosen ratio includes the quantities you know and the quantity you want. For a side answer, include the same units as the given side lengths. For an angle answer, use degrees and state the rounding requested.

Label the triangle before choosing a method.
Use side lengths with Pythagoras and side-to-side ratios.
Check units, angle size, and reasonable side lengths.

Choose a ratio using the sides you need

Ratio	Side comparison	Useful when
Sine	Opposite over hypotenuse	The opposite side and hypotenuse are involved
Cosine	Adjacent over hypotenuse	The adjacent side and hypotenuse are involved
Tangent	Opposite over adjacent	The opposite side and adjacent side are involved

Worked example

Use both side relationships to solve a triangle

A right triangle has legs of

9

cm and

12

cm. Find the hypotenuse and the two acute angles. Round angles to the nearest tenth of a degree.

Find the hypotenuse
The two given sides meet at the right angle, so they are the legs. Let the unknown hypotenuse be $c$ . Apply the Pythagorean theorem, then take the positive square root.
$9^2+12^2=c^2=225,\quad c=15\text{ cm}$
Find one acute angle
Choose the angle opposite the $9$ cm side. Relative to this angle, the $12$ cm side is adjacent and the $15$ cm side is the hypotenuse. Tangent uses opposite divided by adjacent, so use the inverse tangent to find the angle.
$\theta=\tan^{-1}\left(\frac{9}{12}\right)\approx36.9^\circ$
Find the other acute angle
The two acute angles add to $90^\circ$ . Subtract the angle just found from $90^\circ$ to get the remaining angle.
$90^\circ-36.9^\circ=53.1^\circ$

Answer: The hypotenuse is

15

cm. The acute angles are approximately

36.9^\circ

and

53.1^\circ

.

Check: The hypotenuse is longer than either leg, and the two acute angles add to

90^\circ

. Both checks fit a right triangle.

Common mistakes and how to avoid them

Using the selected angle to decide which side is the hypotenuse.

Correction: Find the

90^\circ

angle first. The side across from it is always the hypotenuse.

Using the wrong pair of sides in a trigonometric ratio.

Correction: Mark the chosen acute angle, label opposite and adjacent from that angle, and then select the ratio.

Treating

\sin^{-1}

as the reciprocal of sine.

Correction: In this setting, inverse sine finds an angle from a side ratio. Use the calculator’s inverse function and check that it is in degree mode.

Adding the squares of the legs and reporting that sum as the hypotenuse.

Correction: The sum is the square of the hypotenuse. Take the positive square root to get the side length.

Rounding intermediate values too early.

Correction: Keep calculator values until the final step when practical, then round as requested.

Lesson summary

The hypotenuse lies across from the right angle. Opposite and adjacent depend on the chosen acute angle.
Use the Pythagorean theorem to find a missing side when the other two sides of a right triangle are known.
Use sine, cosine, or tangent to connect a right-triangle angle with two sides.
Use an inverse ratio to find an angle from two known sides. Check the calculator mode and the reasonableness of the result.

Check your understanding

Question 1

A right triangle has legs of

5

cm and

12

cm. What is its hypotenuse?

$13$ cm
$17$ cm
$\sqrt{119}$ cm
$60$ cm

Show answer and explanation

13

cm

The hypotenuse is

\sqrt{5^2+12^2}=\sqrt{169}=13

cm.

Question 2

For an angle

\theta

, the opposite side is

8

and the adjacent side is

6

. Which equation can be used to find

\theta

?

$\theta=\tan^{-1}\left(\frac{8}{6}\right)$
$\theta=\sin^{-1}\left(\frac{8}{6}\right)$
$\theta=\cos^{-1}\left(\frac{8}{6}\right)$
$\theta=\tan^{-1}\left(\frac{6}{8}\right)$

Show answer and explanation

\theta=\tan^{-1}\left(\frac{8}{6}\right)

Tangent compares opposite with adjacent, so the ratio is

\frac{8}{6}

. The inverse tangent returns the angle.

Question 3

A right triangle has hypotenuse

10

cm and an acute angle of

30^\circ

. What is the side opposite the angle?

$5$ cm
$8.7$ cm
$10$ cm
$20$ cm

Show answer and explanation

5

cm

Sine compares opposite with hypotenuse, so the opposite side is

10\sin 30^\circ=5

cm.

Key terms

Right triangle: A triangle with one angle measuring $90^\circ$ .
Hypotenuse: The side across from the right angle. It is the longest side of a right triangle.
Opposite side: The side directly across from a chosen acute angle.
Adjacent side: The side that touches a chosen acute angle and is not the hypotenuse.
Ratio: A comparison of quantities by division.
Inverse ratio: A calculator operation, such as inverse tangent, that finds an angle from a side ratio.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 10 Mathematics (MPM2D), study topic T5. It is a study resource, not an official curriculum publication.

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T5 · Find right-triangle sides and angles using ratios and Pythagoras

Ontario Grade 10 Mathematics

Use Pythagoras and trigonometric ratios to solve missing parts

What you will learn

1. Start with the triangle and its side names

2. Find a missing side with Pythagoras

3. Use sine, cosine, and tangent

4. Guided example and independent practice

Choose a ratio using the sides you need

Use both side relationships to solve a triangle

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through MPM2D

About this lesson