The length of the shadow of a tower standing on level plane is found to be 2x metres longer when the sun' s altitude is 30∘ than when it was 45∘. Prove that the height of tower is x(√3+1) metres.
The length of the shadow of a tower standing on level plane is found to be 2x metres longer when the sun' s altitude is 30∘ than when it was 45∘. Prove that the height of tower is x(√3+1) metres.
The provided information can be displayed diagrammatically as,
Here AB is the tower of height 'h' meter and BC is the length of its shadow, when the sun's attitude is 45° (i.e. when angle of elevation is 45°) and BD is the length of its shadow, when the sun's altitude is 30° (i.e. when angle of elevation is 30°).
Consider right ΔABC,
tan ∠ACB = 
⇒ tan 45° = 
⇒ BC = h
It is given that, BD is 2x meter longer than BC.
So, BD = BC + CD
⇒ BD = BC + 2x
⇒ BD = h + 2x ( 
BC = h)
In rt. ΔADB,
tan ∠D = 
⇒ tan 30° = 
⇒ h + 2x = h 
⇒ 2x = h
– h
⇒ h 
= 2x
Thus, height of the tower is 
meters.
The provided information can be displayed diagrammatically as,
Here AB is the tower of height 'h' meter and BC is the length of its shadow, when the sun's attitude is 45° (i.e. when angle of elevation is 45°) and BD is the length of its shadow, when the sun's altitude is 30° (i.e. when angle of elevation is 30°).
Consider right ΔABC,
tan ∠ACB =
⇒ tan 45° =
⇒ BC = h
It is given that, BD is 2x meter longer than BC.
So, BD = BC + CD
⇒ BD = BC + 2x
⇒ BD = h + 2x ( BC = h)
In rt. ΔADB,
tan ∠D =
⇒ tan 30° =
⇒ h + 2x = h
⇒ 2x = h – h
⇒ h = 2x
Thus, height of the tower is meters.
