Question

Example 6.27 The horizontal distance between two buildings is 140 m. The angle of
depression of the top of the first building when seen from the top of the second building
is 30°. If the height of the first building is 60 m, find the height of the second building.
(13 = 1.732)​

Answer 1

$\huge\frak{\underline{\underline{Question:}}}$

◾️The horizontal distance between two buildings is 140 m. The angle of depression of the top of the first building when seen from the top of the second building is 30°. If the height of the first building is 60 m, find the height of the second building.

(13 = 1.732)

$\huge\frak{\underline{\underline{Answer:}}}$

$\huge{\boxed{\boxed{140.73m}}}$

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$\large\bold{\underline{Explanation:}}$

Let AB be the height of second building and CD be the height of first building.

Given,

BD = AE = 140 m

& AB = DE = 60 m

To Find:

The length of second building

Solutions:

In ΔAEC,

tan 30°= Perpendicular /Base= CE/AE

tan 30 ° = CE/140

1/√3= CE /140

CE= 140 / √3

CE= 140 ×√3/ (√3×√3)

[ Rationalising the denominator]

CE= 140√3 /3

Height of the second building CD= CE+DE

$\implies$ 140√3/3 + 60

$\implies$( 140 × 1.73) /3 + 60

[ √3=1.73]

$\implies$ 242.2/3 + 60

$\implies$ 80.73 + 60

$\implies$ 140.73 m

Height of the second building (CD)=140.73 m

$\rule{400}4$

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