Question

The sides of a $\Delta ABC$ are $AB=\sqrt{13}cm,BC=4\sqrt{3}cm$ and $CA=7cm.$ Then
$sin\theta ,$ where $\theta$ is the smallest angle of the triangle, is equal to

• A. $\frac{\sqrt{3}}{2}$
• B. $\frac{1}{2}$
• C. $\frac{\sqrt{3}-1}{2\sqrt{2}}$
• D. none of these

Answer 1

Answer: (B)

$\frac{1}{2}$

As $\sqrt{13}$ is the smallest side angle opposite to the this will be the smallest angle that is $\theta$.

From cosine rule of triangle $\cos \theta =(a^2+b^2-c^2)/2ab$.

Substitute a =$4\sqrt{3}$$,b=7,c=$ $\sqrt{13}$.Then we get $\cos \theta =\sqrt{3}/2$.

Here we get $\theta$ as 30 degree.

Hence we get sine of the angle c as $1/2$.

Hence option B is the correct answer.