In a $ \triangle A B C $, right angled at $ A $, if $ \tan C=\sqrt{3} $, find the value of $ \sin B \cos C+\cos B \sin C $.

 Given:

In a \( \triangle A B C \), right angled at \( A \), \( \tan C=\sqrt{3} \).

To do:

We have to find the value of \( \sin B \cos C+\cos B \sin C \).

Solution:  

In a triangle $ABC$ right angled at $A$, $tan\ C = \sqrt3$.

We know that,

In a right-angled triangle $ABC$ with right angle at $A$,

By Pythagoras theorem,

$BC^2=AC^2+AB^2$

By trigonometric ratios definitions,

$sin\ B=\frac{Opposite}{Hypotenuse}=\frac{AC}{BC}$

$cos\ B=\frac{Adjacent}{Hypotenuse}=\frac{AB}{BC}$

$sin\ C=\frac{Opposite}{Hypotenuse}=\frac{AB}{BC}$

$cos\ C=\frac{Adjacent}{Hypotenuse}=\frac{AC}{BC}$

Here,

$\tan C=\sqrt{3}=\frac{\sqrt3}{1}$

$BC^2=AC^2+AB^2$

$\Rightarrow BC^2=(1)^2+(\sqrt3)^2$

$\Rightarrow BC^2=1+3$

$\Rightarrow BC=\sqrt{4}=2$

Therefore,

$sin\ B=\frac{AC}{BC}=\frac{1}{2}$

$cos\ B=\frac{AB}{BC}=\frac{\sqrt3}{2}$

$sin\ C=\frac{AB}{BC}=\frac{\sqrt3}{2}$

$cos\ C=\frac{AC}{BC}=\frac{1}{2}$

This implies,

$\sin B \cos C+\cos B \sin C=\frac{1}{2}\times\frac{1}{2}+\frac{\sqrt3}{2}\times\frac{\sqrt3}{2}$

$=\frac{1}{4}+\frac{(\sqrt3)^2}{4}$

$=\frac{1}{4}+\frac{3}{4}$

$=\frac{1+3}{4}$

$=\frac{4}{4}$

$=1$

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

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

817 Views

Kickstart Your Career

Get certified by completing the course

Get Started