Exercises: Solve Right Triangles in Applied Problems

Grade 9·20 problems·~28 min·Common Core Math - HS Geometry·standard·hsg-srt-c-8

Work through problems with immediate feedback

Recall / Warm-Up

A right triangle has one acute angle of 32° and a right angle. What is the measure of the other acute angle?

32°

58°

68°

148°

Which equation correctly applies SOH-CAH-TOA to find the side labeled $x$ in a right triangle where the acute angle is $\theta$ , the hypotenuse is 10, and $x$ is the side opposite $\theta$ ?

$\cos(\theta) = \dfrac{x}{10}$

$\tan(\theta) = \dfrac{x}{10}$

$\sin(\theta) = \dfrac{x}{10}$

$\sin(\theta) = \dfrac{10}{x}$

A right triangle has legs of length 5 and 12. What is the length of the hypotenuse? Round to the nearest tenth if needed.

Fluency Practice

$Right triangle ABC with right angle at C, angle A labeled 40 degrees, side opposite A labeled 7, and hypotenuse labeled with a question mark$

In the right triangle shown, the acute angle at $A$ is 40° and the side opposite $A$ has length 7. Find the length of the hypotenuse. Round to the nearest hundredth.

A right triangle has a hypotenuse of 15 and one acute angle of 60°. Find the length of the side adjacent to the 60° angle. Round to the nearest tenth.

In a right triangle, the side opposite the acute angle $\theta$ has length 9 and the side adjacent to $\theta$ has length 12. Find $\theta$ in degrees. Round to the nearest tenth.

$Right triangle with horizontal leg 6, vertical leg 8, right angle at bottom-right, and the acute angle at the bottom-left labeled with a question mark$

A right triangle has legs of length 6 (horizontal) and 8 (vertical). Find the acute angle at the bottom-left vertex, between the hypotenuse and the horizontal leg. Round to the nearest tenth of a degree.

A right triangle has a hypotenuse of 10 and one acute angle of 35°. Which pair of expressions correctly gives the two legs? (Use $\sin(35\degree) \approx 0.574$ and $\cos(35\degree) \approx 0.819$ .)

Opposite leg $= 10 \times \sin(35\degree) \approx 5.74$ ; adjacent leg $= 10 \times \cos(35\degree) \approx 8.19$

Opposite leg $= \dfrac{10}{\sin(35\degree)} \approx 17.4$ ; adjacent leg $= \dfrac{10}{\cos(35\degree)} \approx 12.2$

Opposite leg $= 10 \times \cos(35\degree) \approx 8.19$ ; adjacent leg $= 10 \times \sin(35\degree) \approx 5.74$

Opposite leg $= 10 \times \tan(35\degree) \approx 7.0$ ; adjacent leg $= 10 \times \sin(35\degree) \approx 5.74$

Varied Practice

$Right triangle with hypotenuse 17, the side opposite theta labeled 8, and the acute angle theta at the bottom-left$

The diagram shows a right triangle with hypotenuse 17 and one acute angle $\theta$ . The side opposite $\theta$ has length 8. Find $\theta$ in degrees. Round to the nearest tenth.

To fully solve the right triangle with hypotenuse 20 and acute angle 50°, complete the following: The side opposite 50° is $20 \times \sin(50\degree) \approx$ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ (to the nearest tenth). The side adjacent to 50° is $20 \times \cos(50\degree) \approx$ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ (to the nearest tenth). The other acute angle is ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ °. Use $\sin(50\degree) \approx 0.766$ and $\cos(50\degree) \approx 0.643$ .

opposite side:

adjacent side:

other acute angle:

A right triangle has two sides known: a leg of length 5 and the hypotenuse of length 13. Which is the most direct way to find the remaining leg?

Use the Pythagorean Theorem: $a^2 + 5^2 = 13^2$

Use inverse sine to find an angle, then use sine to find the leg

Use tangent: $\tan(\theta) = \dfrac{5}{13}$

The leg cannot be found without knowing an acute angle

In a right triangle, you know two sides and need to find one of the acute angles. Which tool should you use?

The Pythagorean Theorem, since it relates all three sides

An inverse trigonometric function ( $\sin^{-1}$ , $\cos^{-1}$ , or $\tan^{-1}$ )

A regular trig ratio (sine, cosine, or tangent), since those connect sides and angles

The angle sum property alone: subtract from 90°

Word Problems

$Side view of a person on the ground looking up at a building at a 35-degree angle of elevation, with horizontal distance 50 m and building height unknown$

A person stands 50 m from the base of a building on flat ground and measures the angle of elevation to the top of the building as 35°. Use $\tan(35\degree) \approx 0.700$ and $\sin(35\degree) \approx 0.574$ .

Find the height of the building. Round to the nearest tenth of a metre.

Find the straight-line distance from the person to the top of the building (the line of sight). Round to the nearest tenth of a metre. Use $\cos(35\degree) \approx 0.819$ .

$A 12-foot ladder leaning against a vertical wall, making a 70-degree angle with the ground, with the height on the wall unknown$

A 12 ft ladder leans against a vertical wall. The base of the ladder makes an angle of 70° with the flat ground.

How high up the wall does the ladder reach? Round to the nearest tenth of a foot. Use $\sin(70\degree) \approx 0.940$ .

A ship leaves port and sails 8 km due north, then turns and sails 6 km due east.

What is the straight-line distance from the port to the ship's final position? Round to the nearest tenth of a kilometre.

Error Analysis

Maya is solving this problem: "In a right triangle, the side opposite the angle $\theta$ is 5 and the side adjacent to $\theta$ is 12. Find $\theta$ ."
Maya writes: $\sin(\theta) = \dfrac{5}{12}$ , so $\theta = \sin^{-1}(0.417) \approx 24.6\degree$ .

What mistake did Maya make?

Maya used sine, but $\frac{\text{opposite}}{\text{adjacent}}$ is the tangent ratio

Maya should have used the Pythagorean Theorem first to find the hypotenuse, then used inverse sine

Maya applied the inverse function incorrectly; she should write $\sin(\theta) = 0.417$ and stop there

Maya's setup is correct; she just rounded incorrectly

Carlos is solving this problem: "From the top of a 40 m cliff, the angle of depression to a boat is 30°. Find the horizontal distance from the cliff base to the boat."
Carlos draws the angle of 30° at the base of the cliff (at the water level), not at the top of the cliff. He then writes: $\tan(30\degree) = \dfrac{d}{40}$ , giving $d = 40 \times \tan(30\degree) \approx 23.1$ m.

What error did Carlos make in his diagram, and what is the correct equation?

Carlos placed the angle correctly but used the wrong trig ratio; he should use cosine

Carlos placed the 30° angle at the bottom of the cliff instead of at the top; the correct equation is $\tan(30\degree) = \dfrac{40}{d}$ , giving $d \approx 69.3$ m

Carlos should not draw a diagram at all and instead use the Pythagorean Theorem

Carlos's diagram is correct; the error is in his arithmetic

Challenge / Extension

$Top-down diagram of a river width problem: points A and B on the near bank 80 m apart, right angle at A, point C directly across from A on the far bank, and angle at B labeled 52 degrees$

A surveyor needs to find the width of a river. She stands at point $A$ on one bank and marks point $C$ directly across the river (so $AC$ is perpendicular to the bank, making a right angle at $A$ ). She then walks 80 m along the bank to point $B$ and sights point $C$ at an angle of 52° from the bank direction (angle $ABC = 52\degree$ ). Find the width of the river ( $AC$ ). Round to the nearest tenth of a metre. Use $\tan(52\degree) \approx 1.280$ .

A ship navigator uses bearings. The navigator says: 'We sailed on bearing N40°E for 100 km.' A crew member interprets this as: 'We sailed 40° from east toward north.' Explain the correct interpretation of N40°E, draw a simple sketch showing the correct direction, and find the northward and eastward components of the 100 km journey. Use $\sin(40\degree) \approx 0.643$ and $\cos(40\degree) \approx 0.766$ .

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