1. In the given figure, ABC is a triangle. The bisectors of internal ∠B and external ∠C interest at D. If ∠BDC = 48°, then what is the value (in degrees) of ∠A?
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By: anil on 05 May 2019 01.48 am
Let $$angle$$ A = $$ heta$$
Using exterior angle property, => $$ heta+2y=2x$$ => $$2(x-y)= heta$$ => $$x-y=frac{ heta}{2}$$ ------------(i) In $$ riangle$$ ABC, => ∠A + ∠B + ∠C = 180 => $$ heta+2y+angle ACB=180^circ$$ => $$angle ACB = 180^circ-2y- heta$$ Thus, ∠DCB = ∠ACB + ∠ACD = $$180^circ-2y- heta+x$$ -----------(ii) Now, in $$ riangle$$ BCD, => ∠DBC + ∠DCB + ∠CDB = 180 => $$y+(180^circ-2y- heta+x)+48^circ=180^circ$$ => $$x-y- heta=-48^circ$$ Substituting value from equation (i), we get : => $$frac{ heta}{2}- heta=-48^circ$$ => $$frac{ heta}{2}=48^circ$$ => $$ heta=48 imes2=96^circ$$ => Ans - (B)
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Using exterior angle property, => $$ heta+2y=2x$$ => $$2(x-y)= heta$$ => $$x-y=frac{ heta}{2}$$ ------------(i) In $$ riangle$$ ABC, => ∠A + ∠B + ∠C = 180 => $$ heta+2y+angle ACB=180^circ$$ => $$angle ACB = 180^circ-2y- heta$$ Thus, ∠DCB = ∠ACB + ∠ACD = $$180^circ-2y- heta+x$$ -----------(ii) Now, in $$ riangle$$ BCD, => ∠DBC + ∠DCB + ∠CDB = 180 => $$y+(180^circ-2y- heta+x)+48^circ=180^circ$$ => $$x-y- heta=-48^circ$$ Substituting value from equation (i), we get : => $$frac{ heta}{2}- heta=-48^circ$$ => $$frac{ heta}{2}=48^circ$$ => $$ heta=48 imes2=96^circ$$ => Ans - (B)