Unit 7 Right Triangles And Trigonometry Homework 1 Answer Key

Unit 7: Right Triangles and Trigonometry – Homework 1 Answer Key

Right triangles are the cornerstone of trigonometry. This article presents a comprehensive answer key for Unit 7, Homework 1, covering all the problems that students typically encounter. But they give us the ability to relate angles to side lengths, a skill that is essential for geometry, physics, engineering, and everyday problem‑solving. Alongside each solution, we explain the underlying concepts, provide step‑by‑step reasoning, and highlight common pitfalls. By the end, you should understand why each answer is correct and how to apply the same logic to similar problems.

Introduction

The first homework set in a unit on right triangles usually tests the following core ideas:

1. Basic definitions – hypotenuse, legs, acute angles.
2. Pythagorean theorem – relationship between the three sides.
3. Trigonometric ratios – sine, cosine, tangent for acute angles.
4. Inverse trigonometric functions – determining angles from ratios.
5. Special right triangles – 30°–60°–90° and 45°–45°–90°.

This answer key walks through each problem, showing the calculations, the reasoning, and the final results. Keep in mind that the values are rounded to two decimal places unless otherwise specified.

Problem 1 – Identify the Hypotenuse

Question
In a right triangle, the legs measure 6 cm and 8 cm. Which side is the hypotenuse?

Solution
The hypotenuse is always the side opposite the right angle and is the longest side.
Using the Pythagorean theorem:

[ c = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ cm} ]

So the hypotenuse is 10 cm.
Common mistake: Forgetting that the hypotenuse is the longest side; some students think the smallest side is the hypotenuse.

Problem 2 – Apply the Pythagorean Theorem

Question
Find the length of the missing side in a right triangle where the hypotenuse is 13 cm and one leg is 5 cm.

Solution
Let the unknown leg be (b).
[ 13^2 = 5^2 + b^2 \ 169 = 25 + b^2 \ b^2 = 144 \ b = 12 \text{ cm} ]

Answer: 12 cm.
Tip: Always square the known sides before adding or subtracting.

Problem 3 – Solve for an Angle Using Sine

Question
A right triangle has legs of 9 cm (adjacent to angle ( \theta )) and 12 cm (opposite to ( \theta )). Find ( \theta ).

Solution
Use the tangent ratio because we have opposite and adjacent sides:

[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{12}{9} = \frac{4}{3} ]

[ \theta = \arctan!\left(\frac{4}{3}\right) \approx 53.13^\circ ]

Answer: ( \theta \approx 53.And 13^\circ ). Common error: Mixing up opposite and adjacent; always check which side is adjacent to the angle in question.

Problem 4 – Find a Missing Side Using Cosine

Question
In a right triangle, the hypotenuse is 10 m and one acute angle is (30^\circ). What is the length of the side adjacent to that angle?

Solution
Cosine relates the adjacent side to the hypotenuse:

[ \cos 30^\circ = \frac{\text{adjacent}}{10} ]

[ \text{adjacent} = 10 \times \cos 30^\circ \approx 10 \times 0.8660 = 8.66 \text{ m} ]

Answer: ( \approx 8.66 \text{ m} ).
Tip: Remember that ( \cos 30^\circ = \sqrt{3}/2 ).

Problem 5 – Special 45°–45°–90° Triangle

Question
A 45°–45°–90° triangle has a hypotenuse of 14√2 cm. What is the length of each leg?

Solution
In a 45°–45°–90° triangle, legs are equal, and the hypotenuse equals leg × √2.

Let leg = (x):

[ x\sqrt{2} = 14\sqrt{2} \ x = 14 \text{ cm} ]

Answer: Each leg is 14 cm.
Common pitfall: Forgetting that the legs are equal in this special triangle.

Problem 6 – Special 30°–60°–90° Triangle

Question
A 30°–60°–90° triangle has a short leg of 5 cm. Find the length of the long leg and the hypotenuse.

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Solution
Properties of a 30°–60°–90° triangle:

• Short leg (opposite 30°) = (x)
• Long leg (opposite 60°) = (x\sqrt{3})
• Hypotenuse = (2x)

Given short leg (x = 5) cm:

[ \text{Long leg} = 5\sqrt{3} \approx 8.66 \text{ cm} ] [ \text{Hypotenuse} = 2 \times 5 = 10 \text{ cm} ]

Answers: Long leg ≈ 8.Consider this: 66 cm; Hypotenuse = 10 cm. Note: Many students mistakenly swap the long and short legs.

Problem 7 – Apply the Law of Sines in a Right Triangle

Question
In a right triangle, angle ( \alpha = 45^\circ ) and side opposite ( \alpha ) is 7 cm. Find the length of the hypotenuse.

Solution
Because the triangle is right, we can use the sine ratio:

[ \sin 45^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{7}{h} ]

[ h = \frac{7}{\sin 45^\circ} = \frac{7}{\sqrt{2}/2} = 7 \times \frac{2}{\sqrt{2}} = 7\sqrt{2} \approx 9.90 \text{ cm} ]

Answer: ( h \approx 9.So naturally, 90 \text{ cm} ). Tip: Recognize that for a 45° angle in a right triangle, the opposite side equals the adjacent side.

Problem 8 – Verify a Right Triangle Using Pythagorean Theorem

Question
Check whether a triangle with sides 9 cm, 12 cm, and 15 cm is right-angled.

Solution
Sort sides so that the largest is the hypotenuse candidate (15 cm):

[ 9^2 + 12^2 = 81 + 144 = 225 \ 15^2 = 225 ]

Since (9^2 + 12^2 = 15^2), the triangle is a right triangle.

Answer: Yes, it is a right triangle.
Common mistake: Treating the smallest side as the hypotenuse. Easy to understand, harder to ignore.

Problem 9 – Find an Angle Using the Tangent Function

Question
A ladder leans against a wall, forming a 75° angle with the ground. The foot of the ladder is 3 m from the wall. How high up the wall does the ladder reach?

Solution
Use the tangent ratio:

[ \tan 75^\circ = \frac{\text{opposite}}{3} \ \text{opposite} = 3 \times \tan 75^\circ \approx 3 \times 3.732 = 11.20 \text{ m} ]

Answer: Approximately 11.20 m high.
Tip: Remember that a large angle close to 90° will produce a large opposite side.

Problem 10 – Determine a Missing Side Using the Sine Function

Question
A right triangle has a 30° angle and the side opposite this angle is 4 m. Find the length of the hypotenuse.

Solution
[ \sin 30^\circ = \frac{4}{h} \ h = \frac{4}{0.5} = 8 \text{ m} ]

Answer: 8 m.
Common error: Confusing the sine ratio with the cosine ratio.

FAQ – Common Questions About Right Triangles

Conclusion

Mastering right triangles and trigonometry requires a blend of memorization (special triangle ratios), procedural skill (applying the Pythagorean theorem and trigonometric ratios), and conceptual understanding (knowing why each step works). Also, remember to keep the definitions fresh, practice the ratios, and always double‑check which side corresponds to which angle. In real terms, by systematically working through the problems above, students build confidence in solving a wide range of related questions. With these tools, you’ll be well‑prepared for the next unit and any real‑world applications that demand trigonometric insight.