Question

The elevation of a tower at a station A due north of it is ${45}^{\circ }$ and at a station B due west of A is ${30}^{\circ }$. If AB = 40 m, find the height of the tower:

• A. 26.26 m
• B. 28.28 m
• C. 38.5 m
• D. None of these

Answer 1

The correct option is B

28.28 m

Let the height of the tower be 'h' m.
Now, since the angle of elevation by A is ${45}^{\circ }$, the distance between station A and the tower is equal to the height of the tower.
Similarly, the distance between station B and the tower is equal to sqrt(3)$\times$h.
Now, the tower, station A and station B together make a right-angled triangle at the ground plane with ${90}^{\circ }$ at A.
=> $3\times h^2={40}^2+h^2$
$h^2=800$
$h=20\times 1.414=28.28m.$