A tower subtends angles `alpha,2alpha,3alpha` respectively, at point `A , B ,a n dC` all lying on a horizontal line through the foot of the tower. Prove that `(A B)/(B C)=1+2cos2alphadot`

Options

A. `(3 sin alpha)/(sin 2 alpha)`
B. `1+2 cos ^2 alpha`
C. `2+cos^3 alpha`
D. `(sin 2 alpha)/(sin alpha )`

Explanation

Let PM the tower .

Given `angleBAP =alpha ,angleCBP= 2alpha and angle MCP = 3alpha `
`therefore angle APB = angle PBC -angleBAP= 2alpha-alpha = alpha`
Similarly, `angle BPC =alpha`
`rArr BP = AB`
By sine formula in triangle BCP , we have
`(BC)/(sin(angleBPC))=(BP)/(sin(angleBCP))`
`rArr (BC)/(sinalpha)=(AB)/(sin(180^@-3alpha))`
`rArr (AB)/(BC)=(sin3alpha)/(sinalpha)=(3sinalpha-4sin^3alpha)/(sinalpha)`
`=3-4sin^2alpha`
` =3-2(1-cos2alpha)`
` =1+2cos2alpha`