Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
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Exercise 22 Page 963

Use the cosine ratio to find m∠ D.

f≈15.6
m∠ D≈ 39.7^(∘)
m∠ E≈ 50.3^(∘)

Practice makes perfect

First, let's draw the measurements from the exercise on a right triangle to visualize the given information. Note that it might not be to scale.

We will find the missing measures one at a time. In this case, this means that we want to find f, m∠ D, and m∠ E.

Hypotenuse

We can find the hypotenuse f using the Pythagorean Theorem. d^2+ e^2= f^2 Let's substitute the known lengths, d = 10 and e= 12, into this equation to find f.

d^2+e^2=f^2
10^2+ 12^2=f^2
â–¼
Solve for f
100 +144+=f^2
244 = f^2
sqrt(244)=f
f = sqrt(244)
f = 15.62049...
f ≈ 15.6

Let's add this value to our diagram.

Angle Measures

We can find m∠ D using the cosine ratio. The cosine of ∠ D is the ratio of the length of the leg adjacent ∠ D to the length of the hypotenuse. cos D=Adjacent/Hypotenuse ⇒ cos D=12/15.6 By the definition of the inverse cosine, the inverse cosine of 1215.6 is the measure of ∠ D. To find it we can use a calculator.

m∠ D=cos ^(-1) 12/15.6
m∠ D =39.71513... ^(∘)
m∠ D≈39.7^(∘)

To find m∠ E, recall that the acute angles of a right triangle are complementary. Therefore, m∠ D and m∠ E add to 90^(∘). m∠ D+m∠ E=90^(∘) Now, we can substitute the approximated measure of ∠ D in our equation and find the measure of ∠ E. 39.7^(∘) +m∠ E≈ 90^(∘) ⇔ m∠ E≈ 50.3^(∘)