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

 Jun 10, 2020
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In triangle(ABC), angle bisectors AX and BY intersect at point I.

--  angle(AIB)  =  109o.

--  angle(ABI) + angle(BAI)  =  180o - 109o  =  71o 

--  therefore, angle(A) + angle(B)  =  2 · 71o  =  142o 

Since angle(A) + angle(B) + angle(C)  =  180o,  angle(C)  =  180o - 142o  =  38o.

 Jun 10, 2020

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