Question

the angle of elevation of the top of the pole from a point on the level ground and 15m away from the pole is 30°. Find the height of the pole.
(*plz ans the ques with a diagram)
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Answer 1

Solution :-

Refer to attachment

Distance between point C from the foot of the pole BC = 15 m

Height of the pole AB = ?

Angle of elevation of the top of the pole from point C ∠ACB = 30°

Consider ΔABC

tan 30° = Opp/Adj = AB/BC = 1/√3

⇒ AB/BC = 1/√3

⇒ AB/15 = 1/√3

⇒ AB = 15/√3 * √3/√3 = 15√3/3 = 5√3

⇒ AB = 5 * 1.732 = 8.66

Therefore height of the pole is 8.66 m

Answer 2

꧁SOLUTION ꧂

✧♥༻༺♥✧

☆GIVEN☆

• The angle of elevation the top of the pole from a point on
• the level and 15m away from the pole is 30

✮LET

• The height of pole be h

☆FIND-The height of the pole

☆According to the above information☆

• In right angle ABC

$⟶tan30= \frac{ab}{bc} \\ \\ ⟶ \frac{1}{ \sqrt{3} } = \frac{h}{15} \\ \\ ⟶h = \frac{15}{ \sqrt{3} } \\ here \: we \: to \: rationalise \\ \\ ⟶ \frac{15 \times \sqrt{3} }{ \sqrt{3} \times \sqrt{3} } \\ \\ ⟶ \frac{15 \times \sqrt{3} }{ \sqrt{9} } \\ \\ ⟶ \frac{15 \sqrt{3} }{3} \\ \\ ⟶5 \sqrt{3}$

⟶Hence:-

• The height of pole =5√3m