Question

from a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20m high building are 45degree and 60 degree respectively. find the height of the tower​

Answer 1

EXPLANATION.

Let the building be = AB

Let the tower be = AC

Angle of elevation from point p to bottom

of tower = 45°

<APB = 45°

Angle of elevation from point p to top of the

tower = 60°

<CPB = 60°

In right angled triangle <APB.

Tan ø = p/b = perpendicular/Base

Tan ø = AB/PB

Tan 45° = AB/PB

= 1 = AB/PB

= AB = PB

= PB = 20 m.

In right angled triangle <CPB.

Tan ø = p/B = perpendicular/Base.

Tan 60° = BC/PB

= √3 = BC/PB

= √3 = BC/20

= BC = 20√3

= AB + AC = 20√3

= 20 + AC = 20√3

= AC = 20√3 - 20

= AC = 20(√3 - 1 )

Height of tower = 20(√3 - 1)m.

Answer 2

✯ Given :-

• A point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively.

✯ To Find :-

• What is the height of the tower.

✯ Solution :-

⋆ Let, DC be the tower and BC be the building.

» Hence,

∠CAB = 45°

∠DAB = 60°

BC = 20 m

➠ In ∆ABC,

⇒ tan45° = $\dfrac{BC}{AB}$

⇒ 1 = $\dfrac{20}{AB}$

➙ AB = 20 m

➠ In ∆ABD,

⇒ tan60° = $\dfrac{BD}{AB}$

⇒ $\sqrt{3}$ = $\dfrac{h + 20}{20}$

➥ h = 20($\sqrt{3}$ - 1) m

$\therefore$ The height of the tower is $\boxed{\bold{\small{20({\sqrt{3} - 1)m}}}}$

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