Question

If ABC are the interior angles of triangle ABC then show that sin B/2 + C/2=A/2

Answer 1

in a triangle, sum of all the interior angles

A + B + C = 180°

⇒ B + C = 180° - A

⇒ (B+C)/2 = (180°-A)/2

⇒ (B+C)/2 = (90°-A/2)

⇒ sin (B+C)/2 = sin (90°-A/2)

⇒ sin (B+C)/2 = cos A/2

Answer 2

Question:

If ABC are the interior angles of triangle ABC then show that $\rm{sin( \frac{B + C}{2} ) = cos \: \frac{A}{2} }$

Solution:

In △ABC we know,

$∠A + ∠B + ∠C = 180 \degree$

∠B+∠C=180° −∠A

Dividing 2 both the sides,

$\rm{⇒ \frac{∠B+∠C}{2} = \frac{180 \degree}{2} − \frac{∠A}{2} }$

Applying sine angle on both the sides,

$\rm{⇒sin( \frac{∠B+∠C}{2} ) = sin}$

$\space$

$\rm{ \implies \: (90 \degree - \frac{ \angle \: A}{2} )}$

$\space$

$\sf{⇒cos( \frac{A}{2} ) }</u></p><p></p><p></p><p><u>[tex] \sf{⇒cos( \frac{A}{2} ) }$

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