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Question 2
ABC is a triangle. The bisectors of the internal angle $$\angle$$B and external $$\angle$$C intersect at D. If $$\angle$$BDC=$$50^{\circ}$$, then $$\angle$$A is
Solution
In $$\triangle$$ BDC,
=> $$y+(180^\circ-2x+x)+50^\circ=180^\circ$$
=> $$y-x+50^\circ=0$$
=> $$y-x=-50^\circ$$
In $$\triangle$$ ABC,
=> $$2y+(180^\circ-2x)+\angle A=180^\circ$$
=> $$2(y-x)+\angle A=0$$
=> $$2(-50^\circ)+\angle A=0$$
=> $$\angle A=100^\circ$$
=> Ans - (A)
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