The side BC of the ΔABC is extended to the point D. If ∠ ACD = 112° and ∠B = 3/4 ∠A, then the value of ∠B is
Given : ∠ ACD = 112° and ∠B = 3/4 ∠A
Let $$\angle$$ A = $$4\theta$$
=> $$\angle$$ B = $$3\theta$$ = ?
Solution : Using exterior angle property
=> ∠A + ∠B = ∠ACD
=> $$4\theta+3\theta=112^\circ$$
=> $$7\theta=112^\circ$$
=> $$\theta=\frac{112}{7}=16^\circ$$
$$\therefore$$ $$\angle$$ B = $$3 \times 16=48^\circ$$
=> Ans - (B)
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