I am stuck with the following problem :
What is the area of an isosceles triangle whose equal sides are $20$ cm and the angle
between them is $30^{\circ}$ ?
It is a nineth standard problem and I can not use trigonometry(I mean I can not use the formula that involves sine ,cosine etc. rule) or integration.
One way to solve it as following :
Consider circumscribed circle and it's radius $R$. By inscribed angle theorem we can have , that $|c|=|R|$, where $c$ is third side of the triangle $a=b=10$. Now using formula $\displaystyle S=\frac{abc}{4R}$, where $S$ is area of triange. So:
$S=\frac{20 \cdot 20 \cdot c}{4R}=\frac{400}{4}=100$
Is there any other simpler way to tackle the problem?
I also don't know how to use the angle as given in the question.
I will be highly obliged if someone gives a detailed clarification to the problem.
Thanks in advance for your time.
