Question

From a top of building 100 m high the angle of depression of two objects arw on.The same side and observed to be 45 and 60 .Find the distance between ghe objects

Answer 1

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Answer 2

42.26 m

Step-by-step explanation:

Let CD be the building such that CD = 100 m.

Let AB be the tower of height h metre. It is given that the angles of depression of the top A and the bottom B of the lower AB are 45° and 60° respectively.

i.e., EAC = 45° and DBC = 60°

Let BD = AE = x In right triangle AEC, we have

tan 45º = CE/AE

1 = 100-h/x

x = 100 - h ...(1) In right triangle BDC, we have

tan 60° = CD/BD

√3=100/x

X=100/√30.. (ii)

Comparing (i) and (ii), we get

100-h = 100/√3

√3(100 - h) = 100

100√3 -√3h = 100

√3h = 100√3-100

h = 100√3-100/√3

h = 100(√3-1)/√3

h = 100(17.32-1/1.732

= 100× 0.732/1.732

= 73.2/1.732

= 42.26 m

Hence,Height of tower (AB) = =

42.26 m.