Question 65

The sides AB and AC of a $$\triangle$$ABC are extended to P and Q respectively. If the bisectors of $$\angle$$PBC and $$\angle$$QCB intersect at O, and $$\angle$$A = $$92^\circ$$, then  $$\angle$$BOC is equal to:

Solution

Now
Angle B +Angle C = 88
Now Angle CBO = (180-B)/2 = 90-B/2 and angle BCO will be 90-C/2
Now in triangle BOC using sum of all angles = 180
we get BOC = 88/2 =44


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