Question

The angles of depression of the top and bottom of a building 50 meters high as observed from the top of a tower are 30 and 60 respectively. Find the height of the tower, and also the horizontal distance between the building and the tower.​

Answer 1

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In ΔBTP,

tan 30° = TP/BP

1/√3 = TP/BP

BP = TP√3

In ΔGTR,

tan 60° = TR/GR

√3 = TR/GR

GR = TR/√3

As BP = GR

TP√3 = TR/√3

3 TP = TP + PR

2 TP = BG

TP = 50/2 m = 25 m

Now, TR = TP + PR

TR = (25 + 50) m

Height of tower = TR = 75 m

Distance between building and tower = GR = TR/√3

GR = 75/√3 m = 25√3 m

Answer 2

$\huge\red{\mid{\underline{\overline{\texttt{Question}}}\mid}}$

The angles of depression of the top and bottom of a building 50 meters high as observed from the top of a tower are 30 and 60 respectively. Find the height of the tower, and also the horizontal distance between the building and

the tower.

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Let AB be the building = 50

CE be the tower = 50+y

Let BC = AD = x=the distance between them

tan 30º =DE/AD

1/V3 = y/x

In right AEBC,

tan60º = CE /BC

V3 = (50 + y)/ X

V3x = 50 + y

V3(V3y) = 50 + y..from (i)

3y - y = 50

2y = 50

y = 25

Height of the tower CE = 50 + y

= 50 + 25

= 75 cm

Now taking the value of y in (i), we get

x = 3 x 25

x = 1.73 x 25

x = 43.25

Required horizontal distance x = 43.25 m

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