Question

50 Points

Answer 1

Answers are in bold.

22. b/c

24. sqrt(17)/17

26. 1/4

28. sqrt(17)/17

30. 1/2

32. sqrt(2)/2

34. sqrt(3)/2

36. Approximately 5.4077 units

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Explanation:

Problem 22

To get the cosine ratio, we divide the adjacent over hypotenuse.

cos(angle) = adjacent/hypotenuse

cos(x) = b/c

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Problem 24

To get the sine ratio, we divide opposite over hypotenuse.

$\sin\left(\text{angle}\right) = \frac{\text{opposite}}{\text{hypotenuse}}\\\\\sin\left(\text{x}\right) = \frac{2}{2\sqrt{17}}\\\\\sin\left(\text{x}\right) = \frac{1}{\sqrt{17}}\\\\\sin\left(\text{x}\right) = \frac{1*\sqrt{17}}{\sqrt{17}*\sqrt{17}}\\\\\sin\left(\text{x}\right) = \boldsymbol{\frac{\sqrt{17}}{17}}$

In the fourth step, I multiplied top and bottom by sqrt(17) to rationalize the denominator.

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Problem 26

tan(angle) = opposite/adjacent

tan(x) = 2/8

tan(x) = 1/4

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Problem 28

cos(y) = sin(x) in this case. The general rule is that sin(A) = cos(B) if and only if A+B = 90.

Therefore the answer is exactly identical to problem 24.

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Problem 30

sin(30) = 1/2 is something you either memorize or look up on a reference chart, or use the unit circle. Alternatively, you can use a 30-60-90 triangle template.

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Problem 32

$\sin\left(45^{\circ}\right) = \boldsymbol{\frac{\sqrt{2}}{2}}$ is something you memorize or have on a reference sheet. You could also use a 45-45-90 triangle.

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Problem 34

$\cos\left(30^{\circ}\right) = \boldsymbol{\frac{\sqrt{3}}{2}}$ is similar to problems 30 and 32 in that you should memorize this or have it on a reference sheet. You could use a 30-60-90 triangle template here.

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Problem 36

Let x be the length of side AC.

Use the tangent ratio to find x.

tan(angle) = opposite/adjacent

tan(B) = AC/AB

tan(31) = x/9

x = 9*tan(31)

x = 5.4077 approximately