c
14. A tree stands vertically on a hill side which makes an angle of 15° with the
horizontal. From a point on the ground 35 m down the hill from the base of the
tree, the angle of elevation of the top of the tree is 60°. Find the height of the
(Ans. 35 V2m)
tree.
Answers
answer
The height of the tree = 35√3 m
Step-by-step explanation:
As shown in the attached figure, let AB be the tree and BD be the hill which makes 15° angle from the horizontal. From D, the angle of elevation of A is 60°
In ΔADB
∠A + ∠ADB + ∠ABD = 180°
⇒ ∠A + (60° - 15°) + (90° + 15°) = 180° (∵ ∠ABD is external angle to the internal angles BDC and ∠BCD
⇒ ∠A + 45° + 105° = 180°
⇒ ∠A = 30°
In ΔADB
By sin rule
\frac{\sin60^\circ}{AB} =\frac{\sin30^\circ}{BD}
AB
sin60
∘
=
BD
sin30
∘
or \frac{\sin60^\circ}{AB} =\frac{\sin30^\circ}{35}
AB
sin60
∘
=
35
sin30
∘
\implies \frac{\sin60^\circ}{AB} =\frac{\sin30^\circ}{35}⟹
AB
sin60
∘
=
35
sin30
∘
\implies 35\sin60^\circ =AB\sin30^\circ⟹35sin60
∘
=ABsin30
∘
\implies 35\times\frac{\sqrt{3}}{2} =AB\times \frac{1}{2}⟹35×
2
3
=AB×
2
1
\implies AB=35\sqrt{3}⟹AB=35
3
Therefore, the height of the tree = 35√3 m