I was asked,
The triangle ABC has for equal parts AB and AC. The midpoint of BC is M, and AM = r. A circular arc with centre A and radius r is drawn and meets the sides AB and AC at the points L and N respectively. Angle CAM = X radians. Given that the area of the sector ALMN is $\frac{1}{4}$ of the triangle ABC, show that $X = \frac{1}{4} \tan X$.
What I did,
$\frac{1}{2} r^2 2X = \frac{1}{4} 2r^2 \tan X$.
From there I conclude $X = \frac{1}{2} \tan X$.
Area of $\triangle BAC$ is $2 \cdot 1/2\cdot r \cdot r \tan X$. Area of the sector $ALMN$ is $(2X)/(2\pi)(\pi r^2)= Xr^2$ . Therefore $Xr^2 =1/4 r^2 \tan X$