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 42 Page 925

Use the sine ratio to find c.

a≈ 26.8
c≈ 28.1
m ∠ A =72.8^(∘)

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 a, c, and m ∠ A.

Side Lengths

We can find c using a sine ratio. The sine of ∠ B is the ratio of the length of the leg opposite ∠ B to the length of the hypotenuse. We know that 8.3 is equal to the length of the leg opposite ∠ B, and that c is equal to the length of the hypotenuse. Therefore, we can write the following equation. Sine=Opposite/Hypotenuse ⇒ sin 17.2 ^(∘) =8.3/c To solve this equation, we will first isolate c. Then, we will have to use the calculator to find the value of sin 17.2 ^(∘).
sin 17.2 ^(∘) =8.3/c
Solve for c
csin 17.2 ^(∘)=8.3
c=8.3/sin 17.2 ^(∘)
c≈8.3/0.295708
c≈ 28.06823
c≈ 28.1
Now, we can find a using the Pythagorean Theorem. a^2 + b^2 = c^2 Let's substitute the known lengths, b = 8.3 and c= 28.1, into this equation to find a.
a^2+b^2=c^2
a^2 + 8.3^2= 28.1^2
Solve for a
a^2+68.89=789.61
a^2=720.72
a=sqrt(720.72)
a=26.84622...
a≈ 26.8

Angle Measures

To find m∠ A, recall that the acute angles of a right triangle are complementary. Therefore, m ∠ A and m ∠ B add up to 90^(∘). m ∠ A + m ∠ B = 90^(∘) Now, we can substitute the measure of ∠ B in our equation and find the measure of ∠ A. m ∠ A+17.2^(∘) = 90^(∘) ⇔ m ∠ A =72.8^(∘)