# How do you find the measures of the angles of a triangle if the measure of one angle is twice the measure of a second angle and the third angle measures 3 times the second angle decreased by 12?

May 9, 2018

$\angle A = {32}^{\circ} , \angle B = {64}^{\circ} , \angle C = {84}^{\circ}$

#### Explanation:

To be certain I understand your question correctly, you have a triangle, where you with "measures" mean number of degrees each angle is.

I.e. let the triangle look something like this:

If $\angle A = {a}^{\circ}$
Then $\angle B = 2 \angle A = 2 {a}^{\circ}$
And $\angle C = 3 \angle A - {12}^{\circ} = 3 {a}^{\circ} - {12}^{\circ}$

If my understanding of your question is correct, we have
$\angle A + \angle B + \angle C = {180}^{\circ}$
Therefore ${a}^{\circ} + 2 {a}^{\circ} + 3 {a}^{\circ} - {12}^{\circ} = {180}^{\circ}$

This gives $6 {a}^{\circ} - {12}^{\circ} = {180}^{\circ}$
Or ${a}^{\circ} = {32}^{\circ}$

The angles in the triangle, therefore, are
I.e. $\angle A = {a}^{\circ} = {32}^{\circ}$
Then $\angle B = 2 \angle A = 2 {a}^{\circ} = {64}^{\circ}$
And $\angle C = 3 \angle A - {12}^{\circ} = {84}^{\circ}$

This gives the following triangle: