Question

Angle bisectors $\overline{AX}$ and $\overline{BY}$ of triangle $ABC$ meet at point $I$. Find $\angle C,$ in degrees, if $\angle AIB = 109^\circ$.

Answer 1

Answer:

<C = $38^{o}$

Step-by-step explanation:

Given that: <AIB = $109^{o}$

<AIB + <BIX = $180^{o}$ (sum of angles on a straight line)

$109^{o}$ + <BIX =  $180^{o}$  

<BIX =  $180^{o}$ - $109^{o}$

<BIX = $71^{o}$

But,

<AIB = <YIX = $109^{o}$ (opposite angle property)

<XIB = <AIY = $71^{o}$  (opposite angle property)

Therefore,

$\frac{A}{2}$ + $\frac{B}{2}$ =  $71^{o}$ (Exterior angle property)

$\frac{A + B}{2}$ =  $71^{o}$

A + B = $142^{o}$

A + B + C =  $180^{o}$ (sum of angles in a triangle)

 $142^{o}$ + C = $180^{o}$

C = $180^{o}$ - $142^{o}$

C = $38^{o}$

Thus, angle C is $38^{o}$.