Question 46

In $$\triangle ABC, BE \perp AC, CD \perp AB$$ and $$BE$$ and $$CD$$ intersect each other at O. The bisectors of $$\angle OBC$$ and $$\angle OCB$$ meet at P. If $$\angle BPC = 148^\circ$$, then what is the measure of $$\angle A$$?

Solution

$$\angle BPC = 148^\circ$$

In triangle BOC-

$$\angle OBC + \angle BCO +\angle BOC = 180$$

$$\angle BOC + 2(\angle PBC + \angle PCB) = 180$$

$$\angle BOC + 2(180 - 148) = 180$$

$$\angle BOC = 180 - 64 = 116$$


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