Question

3. From the top of the tower 60 m high the
angles of depression of the top and bottom of
a vertical lamp post are observed to be 38
and 60° respectively. Find the height of the
lamp post. (tan 38° -0.7813, V3 1.732)
Sol​

Answer 1

Solution :-

Draw a figure with the given information

[ Refer attachment ]

Height of the tower AB = 60 m

Height of the lamp post = CD = ?

Angle of depression from tower A to top of the lamp post C ∠EAC = 38°

Angle of elevation from tower to the bottom of the lamp post D ∠ADB = 60°

Consider ΔACF

∠EAC = ∠ACF [ AE || CF, AC is transversal, alternate angles are equal ]

tan 38° = Opp/Adj = AF/CF

⇒ 0.7813 = AF/CF

⇒ CF = AF/0.7813

⇒ DB = AF/0.7813---eq(1)

[ CF = DB, distance between pole and lamp post ]

Consider the ΔADB

tan 60° = Opp/Adj

⇒ √3 = AB/DB

⇒ √3 = 60/DB

⇒ DB = 60/√3

Rationalizing the denominator

⇒ DB = 60√3/3 = 20√3

⇒ DB = 20 * 1.732

⇒ DB = 34.64

From eq(1)

⇒ AF/0.7813 = 34.64

⇒ AF = 34.64 * 0.7813

⇒ AF = 27.064232

From figure,

FB = AB - AF

FB = CD [ Distance between parallel lines ]

⇒ CD = AB - AF = 60 - 27.064232 = 32.935768 m

Therefore the height of the lamp post is 32.935768 m