JEE Main 2019 — Properties of Triangles Question with Solution

In $∆ ABC,$ angles $A, B, C$ are in $A.P.$ hence $2\angle B=\angle A+\angle C$

Also, by angle sum property $\angle A+\angle B+\angle C=180°$

$\Rightarrow 2\angle B+\angle B=180°$

$\Rightarrow \angle B=60°.$

Now $a:b=1:\sqrt{3}$

By, using sine rule, we have $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

$\Rightarrow \frac{\sin A}{a}=\frac{\sin B}{b}$

$\Rightarrow \frac{\sin A}{\sin \left(60°\right)}=\frac{a}{b}$

$\Rightarrow \sin A=\frac{1}{\sqrt{3 }}·\frac{\sqrt{3 }}{2}=\frac{1}{2}$

$\Rightarrow \angle A=30°$ and hence $\angle C=90°$

Thus, the triangle $ABC$ is a right triangle, with hypotenuse $AB=4 cm.$

Again, by Sine rule, we have $\frac{\sin \left(30°\right)}{a}=\frac{\sin \left(60°\right)}{b}=\frac{\sin \left(90°\right)}{4}$

$\Rightarrow \frac{\left({\displaystyle \frac{1}{2}}\right)}{a}=\frac{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}{b}=\frac{1}{4}$

$\Rightarrow \frac{1}{2a}=\frac{1}{4}$ and $\frac{\sqrt{3}}{2b}=\frac{1}{4}$

$\Rightarrow a=2 cm$ and $b=2\sqrt{3} cm$

Now, the area of $∆ABC=\frac{1}{2}\left(\mathrm{base}\right)\times \left(\mathrm{height}\right)$

$=\frac{1}{2}\left(a·b\right)$

$=\frac{1}{2}\times 2\times 2\sqrt{3}=2\sqrt{3} sq. cm.$