Finding the area of an isosceles triangle with inradius $\sqrt{3}$ and angle $120^\circ$. Different approaches give different results.

Question

I am trying to solve a question which majorly provides these details-

There is an isosceles triangle with the largest angle being $120^\circ$. The radius of its incircle is $\sqrt 3$. The question asks us to find the area of the triangle.

Here's my attempt-

There are many different formulas for the incircle and other relations for the triangle's sides and angles hence using any of them should eventually get the right answer.

We were taught the following formula: the radius of incircle is $$(s-a)\tan(A/2),$$ where $s$ is the semi-perimeter, and $a$ is the side opposite to angle $A$.

I tried using this formula with the side opposite to the $120°$ angle. The $\sqrt 3$ term on both sides gets cancelled leaving me with the result that $$s=a.$$ Further solving with this, I get the result that $$a=b+c$$ where $b$ and $c$ are the equal sides of the isosceles triangle. Hence $$b=c=a/2.$$

On the other hand, I get something different if I try to use the Sine Rule i.e. $$b\sin(A)=a\sin(B).$$ We can compute $B$ to be $30°$ using the Angle Sum Property of a Triangle. I get the result that $$a=b\sqrt 3.$$ Both these results are totally different and can't be true simultaneously. So what fundamental mistake am I making?

Also the first result leads to the conclusion that $$b+c=a,$$ which contradicts the fact that the sum of two sides of a triangle is always greater than the third side.

This question was part of a prestigious examination and the solution I saw simply used the second method i.e. using the sine rule to compute $a$ and then find the semi-perimeter and the area. Then used the result that the radius of the incircle is $$\frac{\text{Area}}{\text{Semi-Perimeter}}$$ to get the final answer according to the Answer Key.

Answer 1

A much easier solution would be to start by recognising that the center of the incircle is the intersection of the internal angle bisectors of the triangle. The triangle is a $120-30-30$ degree triangle, and by rather obvious symmetry, $\angle BEA = 90^{\circ}$.

That gives $AE = \sqrt 3\cot 15^{\circ}$ and therefore the base of the triangle is twice that.

The height $BE = AE \tan 30^{\circ} = \sqrt 3\cot 15^{\circ}\tan 30^{\circ}$

So the area is $\frac 12 (2)\sqrt 3\cot 15^{\circ}\sqrt 3\cot 15^{\circ}\tan 30^{\circ} = 3\cot^2 15^{\circ}\tan 30^{\circ} \approx 24.124$.

Answer 2

Given the value of inradius and all the angles, we can just use this known formula for the area of triangle: \begin{align} S&= r^2\,\cot\tfrac\alpha2\cot\tfrac\beta2\cot\tfrac\gamma2 \tag{1}\label{1} \\ &= 3\,\cot60^\circ\cot^2 15^\circ =12+7\,\sqrt3 \tag{2}\label{2} . \end{align}