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Question:

The angle of elevation of the top of a vertical tower standing on a horizontal plane is observed to be 45° from a point A on the plane. Let B be the point 30 m vertically above the point A. If the angle of elevation of the top of the tower from B be 30°, then the distance (in m) of the foot of the tower from the point A is

15(1+√3)

15(3+√3)

15(3−√3)

15(5−√3)

Solution:

The correct option is B
15(3+√3)
From △ACP, tan45°=x+30/y ⇒ 1=x+30/y ⇒ x+30=y.. (i)
From △BCP, tan30°=x/y ⇒ 1/√3=x/y ⇒ y=√3x.. (ii)
Substituting (ii) in (i), we get
x+30=√3x
30=√3x-x
30=x(√3-1)
x=30/(√3-1)
x=30(√3+1)/(√3-1)(√3+1)
x=30(√3+1)/2
x=15(√3+1)
Therefore, the distance of the foot of the tower from point A is 15(3+√3)m