Question

geometry name date period unit 5 review directions: complete the following sentences with the appropriate key vocabulary. 1. when using the trig function $sinθ$, you must divide the __________ side by the hypotenuse. 2. when using the trig function $cosθ$, you must divide the adjacent side by the ________. 3. when using the trig function $tanθ$, you must divide the opposite side by the ________ side. 4. trig functions $sinθ, cosθ$ and $tanθ$ are only used with ________ triangles. 5. pythagorean theorem is used only with ________ triangles. 6. the hypotenuse is across from the __________. pythagorean theorem solve for the indicated length. 7. $c$ triangle with legs 15 ft and 8 ft 8. $a$ triangle with legs 2.1 cm and 2.9 cm 9. in a right triangle, which side is the hypotenuse and which sides are the legs? trig ratios $sinθ = \frac{opp}{hyp}$ $cosθ = \frac{adj}{hyp}$ $tanθ = \frac{opp}{adj}$ using the triangle below, identify the ratio that represents each trig function. triangle with right angle at b, sides: bc = 36, ab = 15, ac = 39 10. $sinc$ 11. $cosc$ 12. $tanc$ 13. $sina$ 14. $cosa$ 15. $tana$

Explanation:

Step1: Fill trig vocab for $\sin\theta$

$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}$

Step2: Fill trig vocab for $\cos\theta$

$\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}$

Step3: Fill trig vocab for $\tan\theta$

$\tan\theta = \frac{\text{opposite}}{\text{adjacent}}$

Step4: Identify trig function triangle type

Trig ratios apply to right triangles.

Step5: Identify Pythagorean triangle type

Pythagorean Theorem applies to right triangles.

Step6: Identify hypotenuse location

Hypotenuse is across the right angle.

Step7: Solve for $c$ (Pythagorean Theorem)

$c = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17$

Step8: Solve for $a$ (Pythagorean Theorem)

$a = \sqrt{2.9^2 - 2.1^2} = \sqrt{8.41 - 4.41} = \sqrt{4} = 2$

Step9: Define hypotenuse and legs

Hypotenuse: longest side, opposite right angle. Legs: the two shorter sides forming the right angle.

Step10: Calculate $\sin C$

$\sin C = \frac{\text{opposite to } C}{\text{hypotenuse}} = \frac{15}{39} = \frac{5}{13}$

Step11: Calculate $\cos C$

$\cos C = \frac{\text{adjacent to } C}{\text{hypotenuse}} = \frac{36}{39} = \frac{12}{13}$

Step12: Calculate $\tan C$

$\tan C = \frac{\text{opposite to } C}{\text{adjacent to } C} = \frac{15}{36} = \frac{5}{12}$

Step13: Calculate $\sin A$

$\sin A = \frac{\text{opposite to } A}{\text{hypotenuse}} = \frac{36}{39} = \frac{12}{13}$

Step14: Calculate $\cos A$

$\cos A = \frac{\text{adjacent to } A}{\text{hypotenuse}} = \frac{15}{39} = \frac{5}{13}$

Step15: Calculate $\tan A$

$\tan A = \frac{\text{opposite to } A}{\text{adjacent to } A} = \frac{36}{15} = \frac{12}{5}$

Answer:

1. opposite
2. hypotenuse
3. adjacent
4. right
5. right
6. right angle
7. 17 ft
8. 2 cm
9. The hypotenuse is the longest side of the right triangle, opposite the 90° (right) angle. The legs are the two shorter sides that form the right angle.
10. $\frac{5}{13}$
11. $\frac{12}{13}$
12. $\frac{5}{12}$
13. $\frac{12}{13}$
14. $\frac{5}{13}$
15. $\frac{12}{5}$