Question

When one looks at the foot and the top of a tower from the roof of a building, the angles of elevation and depression are 63° and 27° respectively. If the height of the building is 20 metres, find the height of the tower. (tan 63°=2)

Answer 1

||✪✪ QUESTION ✪✪||

When one looks at the foot and the top of a tower from the roof of a building, the angles of elevation and depression are 63° and 27° respectively. If the height of the building is 20 metres, find the height of the tower. (tan 63°=2)

|| ✰✰ ANSWER ✰✰ ||

❁❁ Refer To Image First .. ❁❁

From image we can see That :-

→ BCDE = Rectangle

→ Height of building = DE = 20m

→ BC = DE = 20m (Opp. sides of Rectangle).

→ CD = BE

Now, In ∆CDE , we have ,

→ Tan63° = Perpendicular / Base = ED/CD

→ 2 = 20/CD

→ CD = 10m.

So, BE also = 10m.

Now, In ∆ABE, we Have ,

→ Tan27° = AB/BE

→ (1/2) = AB/10

→ AB = 5m.

So ,,

→ AC = AB+BC

→ AC = 20+5 = 25m

Hence, The height of tower is 25m..

Answer 2

Answer : 25 m

$\rule{100}2$

Given :

• Angles of elevation and depression = 63° and 27° respectively.
• Height of the building = 20 m
• tan 63° = 2

To Find :

• Height of the tower

SolUtion :

From the above info,

tan 63° = BC/CD

⟿ $\sf 2 = \dfrac{20}{CD}$

⟿ $\sf CD =\dfrac{20}{2}$

⟿ $\sf CD = 10$

Moving forward,

➜ cot(90° - ∅) = tan∅

➝  cot(90° - 63)° = tan 63°

➝  cot 27° = tan 63° = 2

➝  cot 27° = BE\AE

∵ BE = CD

➝  $\sf 2 = \dfrac{10}{AE}$

➝  $\sf AE = \dfrac{10}{2}$

➝  $\sf AE = 5$

Also,

Side BC = side DE = 20 m

And AD = AE + DE

$\sf AD = 5\: m\: + 20\:m$

$\sf AD = 25 m$

Hence,

The height of the tower is 25 m.

$\rule{200}2$