Question

The angle of elevation of the top of 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:
(A) 15(1 + √3) (B) 15(3 - √3)
(C) 15(3 + √3) (D) 15(5 - √3)

Answer 1

Answer:

Answer Is in attached image.

Answer 2

The distance between foot of the tower and point A is 15(3+√3).

1) In ∆ACD take tan45° and we got the x=y.

2) Then in ∆DEB take tan30° and here we got the length of BE.

3) It is clearly seen in the diagram that EB and CA are equal.

4) In ΔDCA

tan 45° = P/B= DC /CA

1 = x/y

x = y

• In Δ DEB

tan 30° = DE/ EB

1/ √3 = x-30/y

y = √3(x-30)

x = √3x - 30√3  (x=y)

√3x -x = 30√3

x(√3-1) = 30√3

x = 30√3(√3+1)/ (√3-1)(√3+1)   (rationalise)

  = 30√3(√3+1)/3-1

  = 15(3+√3)

x = y = 15(3 + √3)m