1. In triangle ABC, AB = 12 cm, ∠B = 60°, the perpendicular from A to BC meets it at D. The bisector of ∠ABC meets AD at E. Then E divides AD in the ratio
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By: anil on 05 May 2019 01.58 am
Given : $$angle$$ABC = 60 , AB = 12 cm To find : AE : ED Solution : From $$ riangle$$ABD => $$sin 60 = frac{AD}{BD}$$ => $$frac{sqrt{3}}{2} = frac{AD}{12}$$ => $$AD = 6sqrt{3}$$ cm
Again, => $$cos 60 = frac{BD}{AB}$$ => $$frac{1}{2} = frac{BD}{12}$$ => $$BD = 6$$ cm Also, BF is angle bisector of angle B => $$angle$$EBD = 30 From $$ riangle$$ BDE => $$tan 30 = frac{DE}{BD}$$ => $$frac{1}{sqrt{3}} = frac{DE}{6}$$ => $$DE = 2sqrt{3}$$ cm $$ herefore$$ $$frac{AE}{ED} = frac{AD - ED}{ED}$$ = $$frac{6sqrt{3} - 2sqrt{3}}{2sqrt{3}}$$ = $$frac{4sqrt{3}}{2sqrt{3}} = frac{2}{1}$$ => Required ratio = 2 : 1
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Again, => $$cos 60 = frac{BD}{AB}$$ => $$frac{1}{2} = frac{BD}{12}$$ => $$BD = 6$$ cm Also, BF is angle bisector of angle B => $$angle$$EBD = 30 From $$ riangle$$ BDE => $$tan 30 = frac{DE}{BD}$$ => $$frac{1}{sqrt{3}} = frac{DE}{6}$$ => $$DE = 2sqrt{3}$$ cm $$ herefore$$ $$frac{AE}{ED} = frac{AD - ED}{ED}$$ = $$frac{6sqrt{3} - 2sqrt{3}}{2sqrt{3}}$$ = $$frac{4sqrt{3}}{2sqrt{3}} = frac{2}{1}$$ => Required ratio = 2 : 1