Question

A pole 5m high is fixed on the of a tower.the angle of elevation of the top of pole observed from a point A on the ground is 60degree and the angle of depression of the point A from the top of the is 45degree.findthe height of the tower.(take root3=1.732)

Answer 1

tan 60°=BD/AB
⇒ √3=BD/AB
 ⇒ AB=BD/√3.....(i)

tan 45°=BC/AB
  ⇒ 1=BC/AB
 ⇒  AB=BC.....(ii)
 

From (i) and (ii)
BD/√3=BC
(BC+5)=√3 BC
 5=1.732 BC-BC
5=0.732BC
BC=5/0.732=6.83 m

Answer 2

Let the height of the tower be h 

so $\angle{BAD}=90-45=45$°

the distance of the point A from the foot of tower be b

so so tan 45 = $\frac{h}{b}$

⇒ b = x   {tan 45° = 1}

for ΔABC  $\angle{BAC}=60$°

so tan 60° = $\frac{x+5}{x}$

⇒ √3x = x + 5 

⇒ 0.732x = 5

⇒ x ≈ 6.83 m ANSWER