An isosceles triangle of wood of base 2a and height h is placed with its base on the ground vertex directly above . The triangle faces the sun whose altitude is `30^(@)` . Then the tangent of the angle at the apex of the shadow is

Options

A. `(2ah)/sqrt3`
B. `(2sqrt3ah)/(3h^2-a^2)`
C. `(a^2+h^2)/(2sqrt3)`
D. `(2sqrt3ah)/(3h^2+a^2)`

Answer

B

Explanation

Let ABC be the isosceles triangle with base BC=2a and height AD=h.
Let BCE be the shadow of `triangleABC` such that E is the apex of the shadow.
Clearly, DE is the height of the triangular shadow.
Let `2alpha` be the angle at the apex E of the shadow. Then,
`tan alpha =a/(DE)`

Since the altitude of the sun is `30^@` . Therefore,
`tan 30^@=(AD)/(DE)rArr 1/sqrt3=h/(DE) rArr DE=sqrt3h`
`therefore tan alpha=a/(sqrt3h)`
Now,
`tan 2alpha=(2 tan alpha )/(1-tan^2 alpha ) rArr tan 2alpha =(2a//sqrt3h)/(1-a^2/(3h^2))=(2sqrt3ah)/(3h^2-a^2)`