Question

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 five times the measure of the first angle. The third angle is 14 more than the second. Let x, y, and z represent the measures of the first, second, and third angles, respectively. Find the measures of the three angles.

Answer 1

Answer:

$x=30\\y=68\\z=82$

Step-by-step explanation:

x = measure of the first angle

y = measure of the second angle

z = measure of the third angle

The sum of the measures of the second and third (y+z) is five times the measure of the first angle (=5x)

$y+z=5x$

The third angle is 14 more than the second

$z=y+14$

And remember that the sum of these three angles must be equal to 180.

$x+y+z=180$

Let's take these equations

$y+z=5x\\z=y+14\\x+y+z=180$

If you take a look at the first equation, we have y+z = 5x and we have y+z in the third equation as well, we can replace that....

$x+y+z=180\\x+(y+z)=180\\x+(5x)=180$

Distribute the + sign

$x+5x=180$

Combine like terms;

$6x=180$

Divide by 6.

$x=\frac{180}{6}\\ x=30$

We have now defined that the measure of the first angle is 30º.

Let's take another equation... for example $z=y+14$

I'm going to take this one because if I replace x and z in the third equation, all I'll have left will be y.

$x+y+z=180\\30+y+(y+14)=180$

Distribute the + sign and Combine like terms;

$30+y+y+14=180\\44+2y=180$

Subtract 44 to isolate 2y.

$2y=180-44\\2y=136$

Now divide by 2.

$y=\frac{136}{2}\\ y=68$

We already have the value of x and y. Once again, replacing this in the third equation will leave us with z to solve for.

$x+y+z=180\\30+68+z=180\\98+z=180\\z=180-98\\z=82$