Exercise 32 Page 676

Let's begin by color coding the opposite angles, sides, and the vertices in the given triangle. It will help us use the Law of Sines later.

Let's first find the measure of the third interior angle which is m∠ F, and then the measures of the missing side lengths one at a time.

Finding m ∠ F

To find m ∠ F we can use the Triangle Angle Sum Theorem. This tells us that the measures of the angles in a triangle add up to 180.

108+ 41 + m ∠ F = 180 ⇔ m ∠ F = 31

Finding EF

Note that we know the m ∠ F and the length of the side which is opposite to this angle. We want to find the length of the side EF, that is opposite to ∠ D. Therefore, we can use the Law of Sines! sin D/EF =sin F/DE Let's substitute DE= 7.9, m ∠ F= 31, and m ∠ D= 108 to isolate EF.

sin D/EF =sin F/DE

SubstituteValues

Substitute values

sin 108/EF = sin 31/7.9

▼

Solve for EF

CrossMult

Cross multiply

7.9 * sin 108 = EF * sin 31

DivEqn

.LHS /sin 31.=.RHS /sin 31.

7.9 * sin 108/sin 31 = EF

UseCalc

Use a calculator

14.587943 ... = EF

RearrangeEqn

Rearrange equation

EF = 14.587943 ...

RoundDec

Round to 1 decimal place(s)

EF ≈ 14.6

We found that the length of the side EF is about 14.6 units.

Finding DF

Finally, to find DF we will use the Law of Sines one more time. sin F/DE =sin E/DF Let's substitute DE= 7.9, m ∠ F= 31, and m ∠ E= 41 to isolate DF.

sin F/DE =sin E/DF

SubstituteValues

Substitute values

sin 31/7.9 = sin 41/DF

▼

Solve for DF

CrossMult

Cross multiply

DF * sin 31 = 7.9 * sin 41

DivEqn

.LHS /sin 31.=.RHS /sin 31.

DF = 7.9 * sin 41/sin 31

UseCalc

Use a calculator

DF = 10.063074 ...

RoundDec

Round to 1 decimal place(s)

DF ≈ 10.1

We found that the length of the side DF is about 10.1 units.

Completing the Triangle

With all of the angle measures and side lengths, we can complete our diagram.