Question

The angle of elevation of a building from a distance of 100 m from its foot is 60° . Find the height of the building.

Answer 1

Answer:

$\bigstar{\bold{Height\:=\:100\sqrt{3}\:m}}$

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Distance from the foot (BC) = 100 m
• Angle of elevation = 60°

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Height of the building (AB)

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ Let AB be the height of the building. In ΔABC,

   $\sf{tan\theta = \dfrac{opposite}{adjacent} }$

→ tanθ = AB/BC

→ Substituting value of AB we get,

  tan 60 = AB/100

  √3 = AB/100

   AB = 100√3 m

$\boxed{\bold{Height\:=\:100\sqrt{3} \:m}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

$\sf{tan\theta = \dfrac{opposite}{adjacent} }$

$\sf{sin\theta = \dfrac{opposite}{hypotenuse} }$

$\sf{cos\theta=\dfrac{adjacent}{hypotenuse} }$

Answer 2

✿ AnSwer :

• The height of tower is 173.2 m

━━━━━━━━━━━━━━━

GiVen :

• Distance = 100 m
• Angle of elevation = 60°

To FinD :

• Height of the building = ?

SoluTion :

• Let us assume height of the building be x.

Now, According to the question,

• In △ABC,

$\implies$ tanθ = Perpendicular /Base

$\implies$ tan(60°) = AB / BC

$\implies$ √3 = AB / 100

$\implies$ AB = 100√3

$\implies$ AB = 173.2 m

Therefore, height of the building is 173.2m

━━━━━━━━━━━━━━━━