Question

The angle of elevation of the top q of a vertical tower pq from point x on the ground is 60 degree.At a point y 40m vertically above x the angle of elevation is 45 degree .Find the height of the tower pq and distance xq.

Answer 1

Answer:

Step-by-step explanation:

Given

XY = 40 m

Let the height of the tower AB be ‘h’ m

Hence BC = h – 40

Let AX = CY = x

In right triangle BAX,

tan 60° = (AB/AX)

√3 = h/x

Therefore, x = h/√3 --- (1)

In right triangle BCY,

tan 45° = (BC/CY)

1 = (h – 40)/x

Therefore, x = h – 40

That is [h/√3] = (h – 40) [From (1)]

h = h√3 – 40√3

h(√3 – 1) = 40√3

Therefore, h = 40√3/ (√3 – 1) m

There's an alternate method too:

Let

the height PQ be x. then, as XY is 40, therefore the height from Q to the point

parallel to Y will be x - 40 m. 

 

Now,

x/PX = tan 60 = √3 

 

Also,

(x-40) / PX = tan 45 = 1

 

Solving

for PX, we get, PX = (x-40)

 

Substituting

in eqn 1, we get, x/x-40 = √3

 

Solving

for x, we get x = 40√3 / (√3 - 1) m, which is the required height. 

 

Now,

for XQ, x / XQ = sin 60 = √3/2

Therefore,

solving for XQ, we get XQ = 80/(√3 - 1) m

Hope This Help.