Exercises: Solve Right Triangles in Applied Problems

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

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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 $= 10 \times \sin(35\degree) \approx 5.74$

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