Question

unit 5 test (b) period 1st directions: read through and complete each problem below. show all the necessary work to receive full credit. multiple - choice 1. a right triangle has sides a = 5 and c = 8. what is the length of the missing side? a) 8.1 b) 5.7 c) 6.2 d) 11.0 2. find the hypotenuse of the right triangle. round your answer to the nearest tenth. a) 20.2 m b) 28.0 m c) 410.0 m d) 13.0 m (there is a right triangle image with legs 17 m and 11 m, hypotenuse x) 3. consider the right triangle to the right. what is the perimeter? a) 44 m b) 36 m c) 34 m d) 30 m (there is a right triangle image with one leg 12 m, hypotenuse 15 m, and the other leg b) 4. what is the ratio equivalent to cos b? a) \\(\frac{21}{29}\\) b) \\(\frac{20}{21}\\) c) \\(\frac{21}{20}\\) d) \\(\frac{20}{29}\\) (there is a right triangle image with right angle at c, ac = 21, bc = 20, ab = 29) 5. john is skiing on a mountain with an altitude/height of 1100 feet. the angle of depression is 24°. about how far does john ski down the mountain? a) 1204 ft b) 2470 ft c) 2704 ft d) 447 ft

Explanation:

Problem 1

Step1: Identify hypotenuse and apply Pythagoras

Assume $c$ is hypotenuse. Use $b=\sqrt{c^2-a^2}$.
$\sqrt{8^2-5^2}=\sqrt{64-25}=\sqrt{39}\approx6.2$

Step2: Verify alternative case

If $c$ is leg, hypotenuse would be $\sqrt{5^2+8^2}\approx9.4$, not an option.

Problem 2

Step1: Apply Pythagorean theorem

Use $x=\sqrt{17^2+11^2}$.
$\sqrt{289+121}=\sqrt{410}\approx20.2$

Problem 3

Step1: Find missing side $b$

Use $b=\sqrt{15^2-12^2}$.
$\sqrt{225-144}=\sqrt{81}=9$

Step2: Calculate perimeter

Sum all sides: $12+9+15$.
$12+9+15=36$

Problem 4

Step1: Recall cosine definition

$\cos B=\frac{\text{adjacent to } B}{\text{hypotenuse}}$
$\cos B=\frac{20}{29}$

Problem 5

Step1: Relate angle to triangle

Angle of depression = angle of elevation. Use $\sin(24^\circ)=\frac{1100}{x}$.
$x=\frac{1100}{\sin(24^\circ)}\approx\frac{1100}{0.4067}\approx2704$

Answer:

1. c) 6.2
2. a) 20.2 m
3. b) 36 m
4. d) $\frac{20}{29}$
5. c) 2704 ft