In a $\triangle ABC$ right angled at B, $\angle A = \angle C$. Find the values of$\sin\ A\ cos\ C + \cos\ A\ sin\ C$

Given:

In a $\triangle ABC$ right angled at B, $\angle A = \angle C$.

To do:

We have to find the values of $\sin\ A\ cos\ C + \cos\ A\ sin\ C$.

Solution:  

We know that sum of the angles in a triangle is $180^{\circ}$.

This implies,

$\angle A+\angle B+\angle C=180^{\circ}$

$\angle A+90^{\circ}+\angle A=180^{\circ}$

$2\angle A=180^{\circ}-90^{\circ}$

$\angle A=\frac{90^{\circ}}{2}$

$\angle A=\angle C=45^{\circ}$

Therefore,

$\sin\ A\ cos\ C + \cos\ A\ sin\ C=\sin\ 45^{\circ}\ cos\ 45^{\circ} + \cos\ 45^{\circ}\ sin\ 45^{\circ}$$

$=\frac{1}{\sqrt2}\times\frac{1}{\sqrt2}+\frac{1}{\sqrt2}\times\frac{1}{\sqrt2}$          (Since $\sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt2}$)       

$=\frac{1}{2}+\frac{1}{2}$

$=1$

The value of $\sin\ A\ cos\ C + \cos\ A\ sin\ C$ is $1$. 

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Updated on: 10-Oct-2022

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