Geometry Right Triangle Trigonometry Worksheet Solutions (PDF)

Geometry math worksheet solutions: Right Triangle Trigonometry – Maths PDF

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This right triangles answers for Geometry covers right triangle trigonometry (Common Core G-SRT.C.6, G-SRT.C.7, G-SRT.C.8), with a free PDF to print.

Use this answer key to check the 20 problems on right triangle trigonometry. Every solution shows the formula or theorem used, each step of the work, and the final answer in bold. If an answer differs from yours, compare the steps line by line to find where the work changed.

Geometry · Common Core State Standards: G-SRT.C.6, G-SRT.C.7, G-SRT.C.8

Right triangles answers: exercises 1 to 10

Exercise 1 – Name the ratios

Opposite 3, adjacent 4, hypotenuse 5.

\(\text{sin}\,A = \frac{3}{5}\), \(\text{cos}\,A = \frac{4}{5}\), \(\text{tan}\,A = \frac{3}{4}\).

3/5, 4/5, 3/4

Exercise 2 – Sine of a 5-12-13 triangle

A 5-12-13 triangle: sin A = 5/13, cos A = 12/13, tan A = 5/12.

\(\text{sin}\,A = \frac{5}{13}\)

5/13

Exercise 3 – Tangent

\(\sqrt{8^{2} + 15^{2}} = 17\).

\(\text{tan}\,A = \frac{8}{15}\)

Hypotenuse 17; tan A = 8/15

Exercise 4 – Find a side with sine

Opposite side about 11.5.

\(\text{sin}\,35^{\circ} = \frac{x}{20}\)

\(x = 20 \cdot \text{sin}\,35^{\circ} = 20 \cdot 0.5736 = 11.5\)

x = 11.5

Exercise 5 – Find a side with cosine

\(\text{cos}\,52^{\circ} = \frac{x}{15}\)

\(x = 15 \cdot 0.6157 = 9.2\)

x = 9.2

Exercise 6 – Find a side with tangent

\(\text{tan}\,28^{\circ} = \frac{x}{12}\)

\(x = 12 \cdot 0.5317 = 6.4\)

x = 6.4

Exercise 7 – Find an angle: sine

\(A = \text{sin}^{-1}(0.6) = 36.9^{\circ}\)

36.9°

Exercise 8 – Find an angle from sides

A is about 16.3°.

\(\text{tan}\,A = \frac{7}{24}\)

\(A = \text{tan}^{-1}\left( \frac{7}{24} \right) = 16.3^{\circ}\)

16.3°

Exercise 9 – Complementary angles

Sine of an angle equals the cosine of its complement.

\(x = 90 – 25 = 65\)

x = 65

Exercise 10 – Check complementary

\(\text{sin}\,38^{\circ} = 0.6157\) and \(\text{cos}\,52^{\circ} = 0.6157\).

Both are 0.6157

Exercises 11 and up

Exercise 11 – Solve a right triangle

Solved triangle.

Hypotenuse: \(\sqrt{9^{2} + 12^{2}} = 15\).

Angle opposite 9: \(\text{tan}^{-1}\left( \frac{9}{12} \right) = 36.9^{\circ}\).

Other angle: \(90^{\circ} – 36.9^{\circ} = 53.1^{\circ}\).

Hypotenuse 15; angles 36.9° and 53.1°

Exercise 12 – Solve from hypotenuse and angle

\(90^{\circ} – 40^{\circ} = 50^{\circ}\).

Opposite the 40° angle: \(25 \cdot \text{sin}\,40^{\circ} = 16.1\).

Adjacent: \(25 \cdot \text{cos}\,40^{\circ} = 19.2\).

50°; legs 16.1 and 19.2

Exercise 13 – Solve from a leg and angle

Opposite: \(10 \cdot \text{tan}\,62^{\circ} = 18.8\).

Hypotenuse: \(\frac{10}{\text{cos}\,62^{\circ}} = 21.3\).

Opposite 18.8; hypotenuse 21.3

Exercise 14 – 30-60-90 triangle

Shorter leg: \(14 \div 2 = 7\).

Longer leg: \(7\sqrt{3} = 12.12\). Check with \(\text{sin}\,60^{\circ} = \frac{7\sqrt{3}}{14}\).

7 and 7√3 (about 12.12)

Exercise 15 – 45-45-90 triangle

\(\sqrt{9^{2} + 9^{2}} = 9\sqrt{2} = 12.73\)

9√2, about 12.73

Exercise 16 – Guy wire

Solved guy wire triangle.

Pole height: \(30 \cdot \text{sin}\,62^{\circ} = 26.5\) ft.

Distance: \(30 \cdot \text{cos}\,62^{\circ} = 14.1\) ft.

26.5 ft tall; 14.1 ft from the base

Exercise 17 – Height of a windmill

\(\text{tan}\,41^{\circ} = \frac{h}{80}\)

\(h = 80 \cdot 0.8693 = 69.5\)

69.5 feet

Exercise 18 – Wheelchair ramp

\(\text{tan}\,A = \frac{2.5}{30}\).

\(A = \text{tan}^{-1}(0.0833) = 4.8^{\circ}\)

4.8°

Exercise 19 – Lighthouse

The angle of depression equals the angle of elevation from the boat: \(12^{\circ}\).

\(\text{tan}\,12^{\circ} = \frac{80}{d}\)

\(d = \frac{80}{0.2126} = 376\)

About 376 feet

Exercise 20 – Kite string

\(\text{sin}\,54^{\circ} = \frac{h}{120}\)

\(h = 120 \cdot 0.8090 = 97.1\)

97.1 feet

Key facts behind this right triangles answers

Key facts

sin A = opposite/hypotenuse, cos A = adjacent/hypotenuse, tan A = opposite/adjacent.
sin A = cos (90 degrees minus A) for the acute angles of a right triangle.
Use an inverse function (sin^-1, cos^-1, tan^-1) to find an angle.
Solving a right triangle means finding all sides and acute angles.
Angles of elevation and depression are measured from the horizontal.

How to use this right triangles answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

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