An isosceles triangle of wood is placed in a vertical plane, vertex upwards and faces the sun. If $2a$ be the base of the triangle, $h$ its height and ${30}^{\circ }$ the altitude of the sun, then the tangent of the angle at the apex of the shadow is

$\mathrm{△}XYZ$ is isosceles triangle with $XY=XZ$
$\mathrm{\Delta }YZK$ is the shadow &. Then $\mathrm{\Delta }XIK$ is right angled.

$\mathrm{\Delta }XIK,\cot {30}^{\circ }=\frac{IK}{IX}$
$\sqrt{3}=\frac{IK}{h}\Rightarrow IK=h\sqrt{3}.$

In $\mathrm{\Delta }YIK,\tan \alpha =\frac{a}{h\sqrt{3}}$

Tangent of angle of open of shadow

$$=\tan 2\alpha =\frac{2\tan \alpha }{1-{\tan }^2\alpha }$$

$=2\left(\frac{a}{h\sqrt{3}}\right)/1-\frac{a^2}{3b^2}$
$=\frac{2a\times 3h^2}{h\sqrt{3}\times (3h^2-a^2)}$
$\tan 2\alpha =\frac{2ah\sqrt{3}}{3h^2-a^2}$