Let `alpha` be the solution of `16^(sin^2 theta)+ 16^(cos^2 theta)=10` in `(0,pi//4)` . If the shadow of a vertical pole is `1/sqrt3` of its height , then the altitude of the sun is

Options

A. `alpha`
B. `alpha/2`
C. `2alpha`
D. `alpha/3`

Answer

C

Explanation

We have,
`16^(sin^2 theta)+16^(cos^2 theta)=10`
`rArr 16^(sin^2 theta) + 16^(1-sin^2 theta)=10`
`rArr x+16/x=10` , where x =`16^(sin^2 theta)`
`rArr x^2-10x+16=0 rArr x =2,8`
`therefore 16^(sin^2 theta)=2,8`
`rArr 2^(4 sin^2theta)=2,2^3`
`rArr 4 sin^2 theta=2,3 rArr sin^2 theta=1/2, (sqrt3/2)^2 rArr theta=pi/6,pi/3`

`therefore alpha =pi/6 " " [because pi/3 cancelin (0,pi//4)]`
let the altitude of the sum be `theta`. Then ,
tan `theta =h/(h/sqrt3) rArr tan theta=sqrt3 rArr theta=pi/3 rArr theta =2 alpha`.