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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