Home
Class 10
MATHS
The elevation of a tower at a station A ...

The elevation of a tower at a station A due north of it is `alpha` and at a station B due west of A is `beta`. Prove that the height of the tower is `(ABsinalphasinbeta)/(sqrt(sin^2alpha -sin^2beta))`

Promotional Banner

Similar Questions

Explore conceptually related problems

The elevation of a tower at a station A due north of it is alpha and at a station B due west of A is beta. Prove that the height of the tower is (AB sin alpha sin beta)/(sqrt(sin^(2)alpha-sin^(2)beta))

The elevation of a tower at a Station A due north of it is alpha and at a station B due west of A is beta. Prove that the height of the tower is (AB sin alpha sin beta)/(sqrt(sin^(alpha)-sin^(beta)))

The elevation of a tower at a station A due North of it is alpha and at a station B due West of A is beta .Prove that height of the tower is (AB sin alpha sin beta)/(sin^(2)alpha-sin^(2)beta)

The elevation of a tower at a station A due North of it is alpha and at a station B due West of A is beta . Prove that height of the tower is (A B sinalpha sinbeta)/(sin^2alpha-sin^2beta) .

The elevation of a tower due north of a station A is alpha and at another station B due west of it is beta. If AB=a, the height of the tower is

The elevation of a tower due north of a station A is alpha and at another station B due west of it is beta . If AB =a, the height of the tower is

The angle of elevation of the top of a tower from a point A due south of the tower is alpha and from B due east of tower is beta. If AB=d , show that the height of the tower is (d)/(sqrt(cot^(2)alpha+cot^(2)beta))