1. In ΔABC, ∠A : ∠B : ∠C = 3 : 3 : 4. A line parallel to BC is drawn which touches AB and AC at P and Q respectively. What is the value of ∠AQP - ∠APQ?
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By: anil on 05 May 2019 02.27 am
Given : ∠A : ∠B : ∠C = 3 : 3 : 4 and PQ is parallel to BC To find : ∠AQP - ∠APQ = ? Solution : Let $$angle A=3x$$, $$angle B=3x$$ and $$angle C=4x$$ Thus, in $$ riangle$$ ABC, => $$angle A+angle B+angle C=180^circ$$ => $$3x+3x+4x=180^circ$$ => $$x=frac{180^circ}{10}=18^circ$$ $$ecause$$ PQ $$parallel$$ BC, => $$angle$$ APQ = $$angle$$ B and $$angle$$ AQP = $$angle$$ C (Corresponding angles) $$ herefore$$ $$angle$$ AQP - $$angle$$ APQ = $$4x-3x=x=18^circ$$ => Ans - (B)
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