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McGraw Hill Glencoe Geometry, 2012

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McGraw Hill Glencoe Geometry, 2012 View details

4. Trigonometry

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Exercise 42 Page 575

Recall that the acute angles of a right triangle are complementary. You can use this fact to find the measure of ∠ G.

m ∠ G = 58
GH ≈ 20.8
FH ≈ 17.7

Practice makes perfect

Let's analyze the given right triangle.

We will find the missing measures one at a time. In this case, this means that we want to find m ∠ G, FH, and GH.

Angle Measures

To find m∠ G, recall that the acute angles of a right triangle are complementary. Therefore, m ∠ G and m ∠ H add to 90.

m ∠ G + m ∠ H = 90 Now, we can substitute the given measure of ∠ H in our equation and find the measure of ∠ G. m ∠ G + 32 = 90 ⇔ m ∠ G=58

Side Lengths

We can find GH using a sine ratio.

The sine of ∠ H is the ratio of the length of the leg opposite ∠ H to the length of the hypotenuse. sin ∠ H=opposite/hypotenuse ⇒ sin 32^(∘) =11/GH To find the length of GH, we have to use a calculator.

sin 32^(∘)=11/GH

▼

Solve for GH

LHS * GH=RHS* GH

sin 32^(∘) * GH=11

.LHS /sin 32^(∘).=.RHS /sin 32^(∘).

GH=11/sin 32^(∘)

Use a calculator

GH=20.75787...

Round to 1 decimal place(s)

GH≈ 20.8

Finally, we can find the measure of FH. To do it, we can use the Pythagorean Theorem. (GF)^2 + (FH)^2 = (GH)^2 Let's substitute the known lengths, GF = 11 and GH= 20.8, into this equation to find FH.

(GF)^2 + (FH)^2 = (GH)^2

GF= 11, GH= 20.8

11^2+(FH)^2= ( 20.8)^2

▼

Solve for FH

Calculate power

121+(FH)^2=432.64

LHS-121=RHS-121

(FH)^2= 311.64

sqrt(LHS)=sqrt(RHS)

FH= sqrt(311.64)

Use a calculator

FH = 17.65332...

Round to 1 decimal place(s)

FH≈ 17.7

Subchapter links

Check Your Understanding p.573

Practice and Problem Solving p.573-576

H.O.T. Problems p.576

Standardized Test Practice p.577

Spiral Review p.577

Skills Review p.577