# Ex.8.3 Q6 Introduction to Trigonometry Solution - NCERT Maths Class 10

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## Question

If $$A$$, $$B$$ and $$C$$ are interior angles of a triangle $$ABC$$, then show that

\begin{align} \sin \,\left( {\frac{{B + C}}{2}} \right) = \cos \frac{A}{2}\end{align}

## Text Solution

#### Reasoning:

$$\sin \left( {{{90}^0} - \theta } \right) = \cos \theta$$

#### Steps:

We know that for $$\Delta \,ABC,$$

\begin{align} \angle \,A + \angle \,B + \angle \,C = {180^0}\\ \angle \,B + \angle \,C = {180^0} - \angle \,A \end{align}

On dividing both sides by $$2$$, we get:

\begin{align} \frac{{\angle \,B + \angle \,C}}{2} = \frac{{{{180}^0} - \angle \,A}}{2}\\ \frac{{\angle \,B + \angle \,C}}{2} = {90^0} - \frac{{\angle \,A}}{2} \end{align}

Applying sine angles on both the sides:

\begin{align}\sin \,\left( {\frac{{B + C}}{2}} \right) = \sin \,\left( {{{90}^0} - \frac{A}{2}} \right)\end{align}

Since

\begin{align} \sin \,({90^0} - \theta ) &= \cos \theta \\ ∴ \sin \,\left( {\frac{{B + C}}{2}} \right) &= \cos \,\left( {\frac{A}{2}} \right) \end{align}

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