Question 6
If A, B and C are interior angles of a triangle ABC, then show that $s i n \frac{(B + C)}{2} = c o s \frac{A}{2}$

Open in App

Solution

In a triangle, sum of all the interior angles is equal to 180 $^{\circ}$ .
$\Rightarrow A + B + C = 180^{\circ}$
$\Rightarrow B + C = 180^{\circ} - A$
$\Rightarrow \frac{(B + C)}{2} = \frac{(180^{\circ} - A)}{2}$
$\Rightarrow \frac{(B + C)}{2} = (90^{\circ} - \frac{A}{2})$
$\Rightarrow s i n \frac{(B + C)}{2} = s i n (90^{\circ} - \frac{A}{2})$
$\Rightarrow s i n \frac{(B + C)}{2} = c o s \frac{A}{2}$

Suggest Corrections

Similar questions

sin (\frac{B + C}{2}) = cos \frac{A}{2}

A, B

C

A B C

sin (\frac{B + C}{2}) = cos \frac{A}{2}

s i n \frac{(B + C)}{2} = c o s \frac{A}{2}

Δ

sin \frac{B + C}{2} = cos \frac{A}{2}

If $A, B, C$ are interior angles of $Δ A B C$ then show that, $sin \frac{(B + C)}{2} = cos \frac{A}{2}$

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program