Completing a very difficult triangle

Question

I have an isosceles triangle with the two equal sides of length 'c', and the bottom of length 'a'. Both base angles of the triangle have measures of 'a', in degrees. For example, if 'a' were 50, both these angles would be 50 degrees. What I'm attempting to do is find the missing angle, 'c'. It's measurement in degrees is equal to that of both of the sides of length 'c', much the same way as 'a'. This is what I have so far: 2(a * sin(a)) + a = The perimeter. Also, a * sin(a) / 2 = c. However, this is as far as I can get, as I have no idea what the perimeter is. Does anyone know how to solve a problem like this?

Answer 1

Using sine rule, we get

$$\begin{align*} \frac{a}{\sin c^\circ} =& \frac{c}{\sin a^\circ}\\ a\sin a^{\circ} = &c\sin c^\circ\\ a\sin a^{\circ} =& (180-2a) \sin 2a^\circ\\ a\sin a^{\circ} =&2(180-2a)\sin a^\circ \cos a^\circ\\ a =& 2(180-2a)\cos a^\circ\\ \frac1{\cos a^\circ} =& \frac{360}a-4 \end{align*}$$

For the range of $a\in(0,90)$, the left hand side $\sec a$ is increasing, and the right hand side is decreasing. So the equilateral triangle is the only answer.