Chapter 3.4, Problem 23P

To determine

To find: Measure of each angle in $\Delta \text{ABC}$ .

Answer to Problem 23P

In $\Delta \text{ABC}$ , $\angle \text{A}={\text{84}}^{\text{o}},\angle \text{B}={\text{28}}^{\text{o}},\angle \text{C}={68}^{\circ }$

Explanation of Solution

Given information:

  $\angle \text{A}$ is three times as large as $\angle \text{B}$ .

  $\angle \text{A}$ is ${\text{16}}^{\text{o}}$ larger than $\angle \text{C}$ .

Let

  $\angle \text{A}$ be x .

Then

  $\angle \text{B}=\frac{1}{3}x$

And

  $\angle \text{C}=x-1\text{6}$

We know that

Sum of the angles is

  ${\text{180}}^{\text{o}}$ .

Thus,

  $x+\frac{1}{3}x+x-16=180$

That becomes

  $\frac{7}{3}x-16=180$

Add 16 to both sides:

  $\frac{7}{3}x=196$

Multiply both sides by

  $\frac{3}{7}$ :

  $x=196\left(\frac{3}{7}\right)=84$

Thus,

  $\angle \text{A}={\text{84}}^{\text{o}}$

Then

  $\angle \text{B}=\frac{1}{3}x=\frac{1}{3}\left(84\right)={\text{28}}^{\text{o}}$

And

  $\angle \text{C}=x-\text{16}=84-16={68}^{\circ }$