Exercise 41 Page 676

Let's begin by drawing △ ABC and labeling the lengths of the sides. We will also color code 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∠ G, and then the measures of the missing side lengths one at a time.

Finding m ∠ A

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

26+ 119 + m∠ A = 180 ⇕ m ∠ A = 35

Finding CB

Note that we know the m ∠ B and the length of the side which is opposite to this angle. We want to find the length of the side, a, that is opposite to the ∠ A. Therefore, we can use the Law of Sines. sin B/b =sin A/a Let's substitute b= 15, m ∠ B= 119, and m ∠ A= 35 to isolate a.

sin B/b =sin A/a

SubstituteValues

Substitute values

sin 119/15 = sin 35/a

▼

Solve for a

CrossMult

Cross multiply

a sin 119 = 15 sin 35

DivEqn

.LHS /sin 119.=.RHS /sin 119.

a = 15 sin 35/sin 119

UseCalc

Use a calculator

a = 9.837014...

RoundDec

Round to 1 decimal place(s)

a ≈ 9.8

We found that the length of the side CB is about 9.8 units.

Finding BA

Finally, to find BA we will use the Law of Sines one more time. sin B/b =sin C/c Let's substitute b= 15, m ∠ B= 119, and m ∠ C= 26 to isolate c.

sin B/b =sin C/c

SubstituteValues

Substitute values

sin 119/15 = sin 26/c

▼

Solve for c

CrossMult

Cross multiply

c sin 119 = 15 sin 26

DivEqn

.LHS /sin 119.=.RHS /sin 119.

c = 15sin 26/sin 119

UseCalc

Use a calculator

c = 7.518201 ...

RoundDec

Round to 1 decimal place(s)

c ≈ 7.5

We found that the length of the side BA is about 7.5 units.

Completing the Triangle

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