Question

if a=(b+c)cos theta then show that sin theta =


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Answer 1

$a = (b + c)cos \theta \\ \\ cos \theta = \frac{a}{b + c} \\ \\ {cos}^{2} \theta = \frac{ {a}^{2} }{( {b + c})^{2} } = \frac{ {a}^{2} }{ {b}^{2} + {c}^{2} + 2bc} \\ \\ {sin}^{2} \theta = 1 - \frac{ {a}^{2} }{ {b}^{2} + {c}^{2} + 2bc} \\ \\ {sin}^{2} \theta = \frac{ {b}^{2} + {c}^{2} + 2bc - {a}^{2} }{ {b}^{2} + {c}^{2} + 2bc } \\ \\ {sin}^{2} \theta = \frac{ \frac{ {b}^{2} + {c}^{2} - {a}^{2} }{2bc} + 1}{ \frac{( {b + c})^{2} }{2bc} }$

we know the law of Cosines says that in any triangle, the sum of the squares of two sides minus twice the product of these sides and the cosine of the included angle equals the square of the third side i.e a²=b²+c²-2bc.cosA

$\therefore \: \: {sin}^{2} \theta = \frac{1 + cosA}{ \frac{( {b + c})^{2} }{2bc} } \\ \\ {sin}^{2} \theta = \frac{2 {cos}^{2} \frac{A}{2} }{ \frac{( {b + c})^{2} }{2bc} } \\ \\ {sin}^{2} \theta = \frac{4bc \: {cos}^{2} \frac{A}{2} }{( {b + c})^{2} } \\ \\ sin \theta = \frac{2 \sqrt{bc} }{b + c} cos \frac{A}{2}$