Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
3. Right Triangles and Trigonometric Ratios
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Exercise 50 Page 925

Practice makes perfect
a

We are given a regular pentagon that is inscribed in a circle that has a radius of 10 centimeters.

Notice that we can make triangles with segments of lines between the center and each of the corners of the pentagon. This triangles will be isosceles because the equal sides will be the radius of the circle.

Since we have a regular pentagon, the sides and the angles are equal. Then, we can find the central angle in each triangle by diving 360^(∘) by the number of sides. The number of sides is 5. m∠ C = 360^(∘)/5 = 72^(∘) Therefore, the measure of ∠ C is 72^(∘).

b

We want to find the length of the diagonal PS. To do that, we will start by finding the length of the segment RS. Note that this segment is the opposite leg to ∠ C.

Now, we can recall the trigonometric ratio for sine. sin θ = Opposite/Hypotenuse In this case, the measure of the angle is 72^(∘) and the length of the hypotenuse is 10. We will denote the length of the opposite leg as b. sin θ = Opposite/Hypotenuse ⇓ sin 72^(∘) = b/10 Let's solve this equation to obtain the value of b.

sin 72^(∘) = b/10
10 sin 72^(∘) = b
9.510565 ... = b
9.5 ≈ b
b ≈ 9.5

The length of RS is about 9.5 centimeters. Since the segment RS is the half of the segment PS, we can multiply its length by two to find the length of PS. Let's do it! PS = 2( 9.5) ⇒ PS= 19