1. In Δ ABC, ∠B = 90° and AB : BC = 2 : 1. The value of sin A + cot C is
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By: anil on 05 May 2019 01.54 am
Given : $$angle$$B = 90 and AB : BC = 2 : 1 To find : $$sin A + cot C$$ = ? Solution : Let AB = $$2x$$ and BC = $$x$$ In right $$ riangle$$ABC => AC = $$sqrt{(AB)^2 + (BC)^2}$$ => AC = $$sqrt{4x^2 + x^2} = sqrt{5x^2}$$ => AC = $$sqrt{5}x$$ => $$sin A = frac{BC}{AC}$$ => $$sin A = frac{x}{sqrt{5}x} = frac{1}{sqrt{5}}$$ => $$cot C = frac{BC}{AB}$$ => $$cot C = frac{x}{2x} = frac{1}{2}$$ Now, $$sin A + cot C$$ = $$frac{1}{sqrt{5}} + frac{1}{2}$$ = $$frac{2+sqrt{5}}{2sqrt{5}}$$
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