The angle of elevation of top A of a vertical tower AB from a point P on the ground is 60 degree. At a point Q 40m vertically above P, the angle of elevation is 45 degree. Find the height of the tower AB and the distance PA.
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Answers
Step-by-step explanation:
The vertical tower AB be X m
(A is the top and B is the bottom of the vertical tower)
PQ is another tower/pole/building (P is bottom and Q is top ) and the height is 40 m
The angle of elevator from P to A is 60°
therefore, tan 60° = X/BP. ( ht/base).......(1)
The angle of elevation from point B to Q is 45°
therefore, tan 45%°= 40/BP
As tan 45° = 1 then
1 = 40 / BP
BP = 40 m.
Now X = 40 + Y ..............(2)
Substitute (2) in (1)
tan 60° = 40 + y/40
As (tan 60°}= √3
therefore, y = 40√3 - 40
y = 40(√3-1)
y = 40 x 0.732 = 29.28 m
therefore, the height AB = 40+ 29.28 m
= 69.28 m
The distance between PA = √40^2+ 69.28^2
= √ 1600+ 4799.7184= √6399.7184
= 79.998 m
therefore AB = 69.28 m and AP = 79.998
= 80 m (approx)
Answer:
Step-by-step explanation:
Find = AB ,AP
Solution= in triangle BPQ
tan45 =QP/PB
1= 40/BP
BP=40m
In triangle ABP
tan60=AB/PB
√3=AB/40
AN=40√3 m
Cos 60= BP/AP
1/2=40/AP
AP=80 m