The hypotenuse of a right triangle is equal to $c$, one of the acute angles being 30°. Find the radius of the circle with centre at the vertex of the angle of 30° which separates the triangle into two equivalent parts.
My attempt: Let angle $B$ be 90°, $A$ be 30°.{WLOG}. Then, angle $C=60°$.
Now, let the circle have its centre at $C$, with radius such that it passes through $B$. Let it intersect $AC$ at $X$, such that $BX$ is perpendicular to $AC$. This gives two similar triangles, $ABX$ and $BXC$.
Thereafter, I don't know what to do. The ratios seem to be very confusing, but I guess we have to find length of $BC$??
The solution given is:
$$\frac c 2 √({\frac {3√3} π})$$
I'm stuck at 0 progress... Please help. Much obliged. (Sorry for bad formatting.)
The area of the triangle is $\large{A=\frac{ab}{2}=\frac{c \sin 30° \cdot c \cos 30°}{2}=\frac{c^2\sqrt{3}}{8}}$. Let $r$ be a radius of the circle. The area of circular sector is $1 \over 12$ of the whole circle $(30:360)$ or $\frac{\pi r^2}{12}$. So we have this equation $$ \frac{c^2\sqrt{3}}{8}-\frac{\pi r^2}{12}=\frac{\pi r^2}{12}$$ Solving this will give you the answer.