# A bird flying at height can see the top of the two buildings of height 534 m and 300 m.

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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.

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by (45.9k points) (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.