Pearson Algebra 2 Common Core, 2011
PA
Pearson Algebra 2 Common Core, 2011 View details
Chapter Test
Continue to next subchapter

Exercise 24 Page 963

Use the cosine ratio to find m∠ D.

f≈6.1
m∠ D≈ 64.8^(∘)
m∠ E≈ 25.2^(∘)

Practice makes perfect

First, let's draw the measurements from the exercise on a right triangle to visualize the given information.

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 by using the Pythagorean Theorem. d^2+ e^2= f^2 Let's substitute the known lengths, d = 5.5 and e= 2.6, into this equation to find f.

d^2+e^2=f^2
5.5^2+ 2.6^2=f^2
â–¼
Solve for f
30.25 + 6.76 = f^2
37.01= f^2
sqrt(37.01)=f
f = sqrt(37.01)
f = 6.08358...
f ≈ 6.1

Let's add this value to our diagram.

Angle Measures

We can find m∠ D by 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=2.6/6.1 By the definition of the inverse cosine, the inverse cosine of 2.66.1 is the measure of ∠ D. To find it, we can use a calculator.

m∠ D=cos ^(-1) 2.6/6.1
m∠ D =64.77148... ^(∘)
m∠ D≈ 64.8^(∘)

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. 64.8^(∘) +m∠ E≈ 90^(∘) ⇔ m∠ E≈ 25.2^(∘)