Let the radius of the circle be r and let the perpendicular height of the triangle be h.

$\tan x^{\circ}= h/r$

Now, the area of the semicircle = ${1\over2}\pi r^2$ and the area of the triangle = ${1\over2}\times 2r \times h$

Which gives $r h$ = ${1\over2}\pi r^2$, so ${h\over r} = {\pi \over 2}$

*This problem is taken from the UKMT Mathematical Challenges.*