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McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

2. Surface Areas of Prisms and Cylinders

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Exercise 54 Page 821

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

x≈ 20.5

Practice makes perfect

We are given the length of one leg and the measure of an acute angle of a right triangle. We want to find the length of the hypotenuse.

Note that the given side is opposite the given angle, and the side we want to find is the hypotenuse. Therefore, we will use the sine ratio. sin θ = opposite/hypotenuse In our triangle, we have that θ =23^(∘) and the length of the opposite leg is 8. We want to find the length of the hypotenuse.

sin θ = opposite/hypotenuse

SubstituteValues

Substitute values

sin 23^(∘) = 8/x

▼

Solve for x

LHS * x=RHS* x

sin 23^(∘) * x = 8

.LHS /sin 23^(∘).=.RHS /sin 23^(∘).

x = 8/sin 23^(∘)

Use a calculator

x=20.474437...

Round to 1 decimal place(s)

x≈ 20.5

Subchapter links

Exercises p.817-821