McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
1. Trigonometric Functions in Right Triangles
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Exercise 38 Page 796

Determine the trigonometric ratio to use according to the given information and the unknown.

x = 93.7, y = 60.2

Practice makes perfect

We are given the length of one leg and the measure of an acute angle of a right triangle. We will find values of x and y one at a time.

Value of x

x is a length of the hypotenuse.

Note that the given side is opposite to the given angle, and the side we want to find is the hypotenuse of the triangle. Therefore, we will use the sine ratio. sin θ = opposite/hypotenuse In our triangle, we have that θ = 50^(∘) and the length of the opposite side is 71.8. We want to find the length of the hypotenuse.

sin θ = oposite/hypotenuse
sin 50^(∘) = 71.8/x
â–¼
Solve for x
xsin 50^(∘) = 71.8
x = 71.8/sin 50^(∘)
x=93.728243...
x≈ 93.7

Value of y

y is a length of the side adjacent to the given angle.

As we can see, the given side is opposite to the given angle, and the side we want to find is adjacent to the given angle. This means that we will use the tangent ratio. tan θ = opposite/adjacent Similar as before, we have that θ = 50^(∘) and the length of the opposite side is 71.8. Let's find the length of the leg adjacent to the angle.

tan θ = opposite/adjacent
tan 50^(∘) = 71.8/y
â–¼
Solve for y
y tan 50^(∘) = 71.8
y = 71.8/tan 50^(∘)
y=60.247353...
y≈ 60.2