Q9 of 69 Page 9

In Δ ABC, if ∠A = 100° AD bisects ∠A and AD⊥BC. Then, ∠B =

Given,

AD perpendicular to BC


∠A = 100o


In ,


∠ADB + ∠B + ∠DAC = 180o


90o + ∠B + ∠A = 180o


∠B + * 100o = 180o – 90o


∠B + 50o = 90o


∠B = 40o

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