In triangle $A B C$, the measure of angle $B$ is $68^{\circ}$ more than twice the measure of angle $A$. The measure of angle $\mathrm{C}$ is $64^{\circ}$ more than that of angle $\mathrm{A}$. Find the angle measures.
A) $12^{\circ}, 92^{\circ}, 64^{\circ}$
B) $12^{\circ}, 92^{\circ}, 76^{\circ}$
C) $15^{\circ}, 98^{\circ}, 67^{\circ}$
D) $8^{\circ}, 84^{\circ}, 88^{\circ}$

Question

Quizplus · Accepted Answer

The sum of angles in a triangle is $180^{\circ}$. Let angle $A$ be $x$. Then, angle $B = 2x + 68^{\circ}$ and angle $C = x + 64^{\circ}$. Solving $x + (2x + 68^{\circ}) + (x + 64^{\circ}) = 180^{\circ}$ gives $x = 12^{\circ}$, so angles are $12^{\circ}, 92^{\circ}, 76^{\circ}$.

In Triangle ABCA B CABC , the Measure of Angle BBB Is 68∘68^{\circ}68∘

In Triangle $A B C$ , the Measure of Angle $B$ Is $68^{\circ}$