Q 191 - Intermediate Algebra

Question

In the following exercises, solve the given problem. 

The sum of the measures of the angles of a triangle is $180$. The sum of the measures of the second and third angles is three the measure of the first angle. The third angle is fifteen more than the second. Find the measures of the three angles. 

Step-by-Step Solution

Answer

The measures of the three angles are $45,60,75$

Step 1. Identify and name what we are to find

We need to find the measures of the three angles of the triangle.

Let the measure of the first angle be $x$, the measure of the second angle be $y$, and the measure of the third angle be $z$

Step 2. Form the equations

As the sum of the three angles is $180$, so the first equation is

$x+y+z=180 ...\left(1\right)$

The sum of the measures of the second and third angles is three times the measure of the first angle, so the second equation is

$y+z=3x ...\left(2\right)$

The third angle is fifteen more than the second, so the third equation is 

$z=y+15 ...\left(3\right)$

Step 3. Substitute the value of z in the first two equations

Using the third equation, substitute $y+15$ for $z$ in the third equation

$\begin{array}{rcl}x+y+z& =& 180\\ x+y+y+15& =& 180\\ x+2y+15-15& =& 180-15\\ x+2y& =& 165 ...\left(4\right)\end{array}$

Again, substitute $y+15$ for $z$ in the third equation

$\begin{array}{rcl}y+z& =& 3x\\ y+y+15& =& 3x\\ 2y+15-15-3x& =& 3x-15-3x\\ -3x+2y& =& -15 ...\left(5\right)\end{array}$

Step 4. Solve by elimination

Subtract the fifth equation from the fourth equation

$\begin{array}{rcl}x+2y-(-3x+2y)& =& 165-(-15)\\ x+2y+3x-2y& =& 165+15\\ 4x& =& 180\\ x& =& 45\end{array}$

Step 5. Find the values of y and z

Substitute $45$ for $x$ in the fourth equation

$\begin{array}{rcl}x+2y& =& 165\\ 45+2y& =& 165\\ 45+2y-45& =& 165-45\\ 2y& =& 120\\ y& =& 60\end{array}$

Substitute $60$ for $y$ in the third equation

$$\begin{gathered} z=15+y \\ z=15+60 \\ z=75 \end{gathered}$$

So the measure of the three angles are $45,60,75$