Question

$\huge{\textbf{{{\color{navy}{Que}}{\purple{ѕt}}{\pink{ion♤࿐}}{\color{pink}{:}}}}}$

If A, B and C are interior angles of a triangle ABC, then show that sin [(B + C)/2] = cos A/2.

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Answer 1

$\bigstar \: \underline{\underline \bold{prove : }}$

$\leadsto \sf \red{ \sin ( \frac{B+ C}{2} ) = \cos( \frac{A}{2} ) }$

$\underline{ \underline{ \bigstar \: \bold{proof : }}}$

$\odot \sf{ \: as \: we \: know : - }$

$\underbrace\bold{ : \longmapsto \angle \: a + \angle \: b + \angle \: c = 180 \degree}$

$\pmb{( \angle \: e\: sum \: property)}$

$\downarrow \purple{ \pmb{\:dividing \: both \: the \: sides \: by \: 2}} \: \downarrow$

$\bold{ \hookrightarrow \frac{\angle \: a} {2} + } \bold{\frac{ \angle \: b}{2} + } \bold{\frac{ \angle \: c}{2} = } \bold{\frac{180 \degree}{2} }$

$\bold{ \hookrightarrow \frac{\angle \: a} {2} + } \bold{\frac{ \angle \: b}{2} + } \bold{\frac{ \angle \: c}{2} = \: } \bold {{90 \degree}}$

$\hookrightarrow\bold{\frac{ \angle \: b}{2} + } \bold{\frac{ \angle \: c}{2} = \: } \bold{{90 \degree} \bold{- \frac{ \angle \: a}{2} }}$

$\bigstar \: \pmb{\purple{on \: taking \: 'sin' \: in \: both \: the \: sides : - }}$

$\sf\hookrightarrow \: \sin( \frac{\angle \: b + \angle \: c}{2}) = \sin(90 \degree - \frac{ \angle \: a}{2} )$

$\sf\hookrightarrow \: \sin( \frac{\angle \: b + \angle \: c}{2}) = \cos({ \frac{ \angle \: a}{2} })$

$\sf\hookrightarrow \: \sin( \frac{\:B + \: C}{2}) = \cos\frac{A} {2}$

$\overline {\underline{\fbox{| hence \: proved |}}}$

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Answer 2

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➪Given△ABC

➪We know that sum of three angles of a triangle is 180

➪Hence ∠A+∠B+∠C=180°

➪or A+B+C=180°

➪B+C=180° −A

➪Multiply both sides by 1/2

➪1/2 (B+C) = 1/2 (180° − A)

➪1/2 (B + C) = 90° - A/2...(1)

➪Now 1/2 (B+C)

➪Taking sine of this angle

➪sin (B + C /2)

➪[B + C /2 = 90° - A/2]

➪sin(90°-A/2)

➪cos A/2 [sin(90° - θ ) = cos θ]

➪Hence sin( B + C/2 )=cos A/2 proved

❝Hope It Helps❞

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