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In any ` DeltaABC`, prove that:`(sinB)/(sinC)=(c-acosB)/(b-acosC)` |
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Answer» `R.H.S. = (c-acosB)/(b-acosC)` From projection formulas, `c = acosB + bcosA` `b = acosC+c cosA` `:.(c-acosB)/(b-acosC) = (acosB + bcosA-acosB)/(acosC+c cosA-acosC)` `=(bcosA)/(c cosA)` `=b/c` `L.H.S. = sin B/sinC` From Sine formulas,`sinA = ka, sinB = kb, sin C = kc` `:. sin B/sinC = (kb)/(kc) = b/c` `:. L.H.S. = R.H.S.` |
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