Question

the angle of the elevation of the top of a tower standing on a horizontal plane from a point A is alpha .After walking a distance d towards the door of the tower, the angle of elevation is found to be beeta then the height of the tower is
(1)d/tan alpha-tan beeta

(2)d( cot alpha- cot beeta)
(3)d/ cot alpha- cot beeta
(4) d( tan alpha- tan beeta)​

Answer 1

Step-by-step explanation:

Let AB be the tower and C is a point such that the angle of elevation of A is α.

After walking towards the foot B of the tower, at D the angle of elevation is β.

Let h be the height of the tower and DB = x Now in ΔACB,

Height and Distance mcq solution image

tanθ=PerpendicularBase=ABCBtanα=hd+x⇒d+x=htanα⇒d+x=hcotα⇒x=hcotα−d.......(i)Similarly in right ΔADB,tanβ=hx⇒x=htanβ⇒x=hcotβ.........(ii)From(i)and(ii)hcotα−d=hcotβ⇒hcotα−hcotβ=d⇒h(cotα−cotβ)=d⇒h=dcotα−cotβ

∴ Height of the tower =dcotα−cotβ