Question

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${\LARGE{\sf Question}}$
The height of a building is 10m.From the top of the building, there is a pole Situated over a Flat Surface. The angle of elevation and the angle of depression of the top and bottom of the pole are 60° and 45° respectively.
(1)-Find the height of the pole.
(2)-Find the distance between Pole and the building.
${\huge{\sf Note:-}$Use trigonometric ovations to Solve this .
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Answer 1

Given :- (From image.)

• AB = Building = 10m .
• EC = ED + DC = (x + 10) m = Pole.
• ∠EAD = 60°
• ∠ACB = 45°

To Find :-

• EC = ?
• BC = ?

Solution :-

in Right ∆ABC , we have,

→ Tan 45° = Perpendicular / Base = AB/BC

→ Tan 45° = 10/BC

→ 1 = 10/BC

→ BC = 10 m. (Ans.)

Now, since,

→ BC = AD = 10m.

then, in right ∆ADE ,

→ Tan 60° = Perpendicular / Base = ED/AD

→ Tan 60° = x/10

→ √3 = (x/10)

→ x = 10√3 m.

therefore,

→ EC = ED + DC

→ EC = 10√3 + 10

→ EC = 10(√3 + 1) m. (Ans.)

Hence, Height of Pole is 10(√3 + 1)m and the distance between Pole and the building is 10m.

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Answer 2

Given:- (From Image)

• AB = Building = 10m
• EC = ED + DC = (x + 10)m = Pole
• $\angle$ EAD = 60°
• $\angle$ ACB = 45°

⠀

⠀

To find :-

• EC = ?
• BC = ?

⠀

⠀

Solution:

➖In Right ΔABC, we have,➖

➡Tan 45° = Perpendicular/Base = AB/BC

➡Tan 45° = 10/BC

➡1 = 10/BC

➡BC = 10m. (Ans.)

⠀

➖Now, Since,➖

➡BC = AD = 10m.

⠀

➖Then, In Right ΔADE,➖

➡Tan 60° = Perpendicular/Base = ED/AD

➡Tan 60° = x/10

➡√3 = (х/10)

➡х = 10√3 m

⠀

Therefore,

➡EC = ED + DC

➡EC = 10√3 + 10

➡EC = 10(√3 + 1) m.

⠀

Hence, the height of Pole is 10(√3 + 1) meter and the distance between the pole and the building is 10 meter.