# 1969 AHSME Problems/Problem 8

## Problem

Triangle $ABC$ is inscribed in a circle. The measure of the non-overlapping minor arcs $AB$, $BC$ and $CA$ are, respectively, $x+75^{\circ} , 2x+25^{\circ},3x-22^{\circ}$. Then one interior angle of the triangle is: $\text{(A) } 57\tfrac{1}{2}^{\circ}\quad \text{(B) } 59^{\circ}\quad \text{(C) } 60^{\circ}\quad \text{(D) } 61^{\circ}\quad \text{(E) } 122^{\circ}$

## Solution $[asy] draw(circle((0,0),65)); draw((25,60)--(39,-52)--(-52,-39)--(25,60)); dot((25,60)); dot((39,-52)); dot((-52,-39)); dot((0,0)); draw((0,0)--(-52,-39)); draw((0,0)--(39,-52)); draw((0,0)--(25,60)); label("A",(-52,-39),SW); label("B",(25,60),NE); label("C",(39,-52),SE); [/asy]$ Because the triangle is inscribed, the sum of the minor arcs equals $360^\circ$. Thus, $$x+75+2x+25+3x-22=360$$ $$6x+78=360$$ Solving this yields $x = 47$, so the inscribed angles are $122^\circ$, $99^\circ$, and $119^\circ$. Noting that an angle of $\triangle ABC$ is half of its corresponding inscribed angle, so the angles of $\triangle ABC$ are $59.5^\circ$, $49.5^\circ$, and $\boxed{\textbf{(D) } 61^\circ}$.

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