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'Ice Cream Tangent' printed from https://nrich.maths.org/
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 ${h\over r} = {\pi \over
2}$