The angle elevation of the top Q of a vertical tower PQ from a point Xon the ground is 60 degree . at a point Y, 40 m verticaaly above X, the angle of elevation is 45 degree . find the height of the tower PQ and the distance XY.
Answers
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 Helps :)
Answer:h=(40√3)/(√3-1)
Step-by-step explanation: follow the below attachment.
HOPE THIS HELPS YOU.:-)
