Question

The angles of depression of the top and bottom of a 50 m high building from the top of a tower are 45° and 60° respectively. Find the height of the tower and the horizontal distance between the tower and the building.

(Use √3 = 1.73)

Comments

It is right answer

Answer 1

Height of the tower = 118.25 m

Distance between the tower and the building = 68.25 m 

Answer 2

h = 75 + 25 √3

   = 118.25 m

Comments

It’s absolutely crrct ✌️

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The answer correct

Answer 3

Let the height of the tower AB be h m and the horizontal distance between the tower and the building BC be x m.

So,

AE = (h−50) m

In ∆AED,

\(\tan45° = \frac{AE}{ED}\)

⇒ \(1 = \frac{h - 50}x\)

⇒ x = h − 50   .....(1)

In ∆ABC,

\(\tan60° = \frac{AB}{BC}\)

⇒ \(\sqrt 3 = \frac hx\)

⇒ \(x = {\sqrt 3} = h\)   ......(2)

Using (1) and (2), we get

\(x = \sqrt 3x - 50\) 

⇒ \(x = (\sqrt 3 - 1) = 50\)

⇒ \(x = \frac{50(\sqrt 3 + 1)}2\)

\(= 25 \times 2.73\)

\(= 68.25\)

Substituting the value of x in (1), we get

68.25 = h − 50

⇒ h = 68.25 + 50

⇒ h = 118.25 m

Hence, the height of tower is 118.25 m and the horizontal distance between the tower and the building is 68.25 m.