Question

A bird flying at height can see the top of the two buildings of height 534 m and 300 m. The angles of depression from bird, to the top of first and second buildings are 30° and 60° respectively. If the distance between the two buildings is 142 m, and the bird is vertically above the midpoint of the distance between the two buildings, answer the following questions.

i) Draw a labelled figure on the basis of the given information and find the
approximate height of the bird from ground.

ii) Find the distance between bird and top of building I.

Answer 1

(i) Let both buildings are AB and CD respectively.

Now, in right ΔBME, ∠B = 30°,

tan 30° = ME/BM

⇒ ME = BM x tan30°

 = 71/√3 M (\(\because tan30^\circ = \frac1{\sqrt3}\,and\, BM = 71m\))

Now, EF = FM + ME

 = (534 + 71/√3) = (534 + 40.99)m = 574.99 m

Hence, the approximate height of the bird from ground is 574.99 ≅ 575 m.

(ii) BE = \(\sqrt{{BM^2}+{ME^2}}\)

(By using pythagoras theorem in right ΔBME)

 = \(\sqrt{71^2+(\frac{71}{\sqrt3})^2}\) (\(\because\) BM = 71m and ME = 71/3  m)

= \(\sqrt{{71^2}+\frac{71^2}2}\) = 71\(\sqrt{1+\frac13}\) = 71\(\sqrt{\frac43}\) = \(\frac{71\times2}{\sqrt3}\) = \(\frac{142}{\sqrt3}m\) 

Hence, the distance between Bird(E) and top of building I(B) is BE = \(\frac{142}{\sqrt3}m\) \(\approx\) 81.98 m.