The angle of elevation of the top of a tower from a certain point is 30°. If the observer moves 40 m towards the tower, the angle of elevation of the top of the tower increases by 15°. The height of the tower is:
64.2 m
2 m
52.2 m
54.6 m
Explanation:
Let DC be the tower and A and B be the positions of the observer such that AB = 40 m
We have $\angle$DAC = 30°, $\angle$DBC = 45°
Let DC = h
tan30°=$\dfrac{DC}{AC}$
=>$\dfrac{1}{\sqrt{3}}=\dfrac{h}{AC}$
=>AC = h√3 ⋯(1)
tan45°=$\dfrac{DC}{BC}$
=>1=$\dfrac{h}{BC}$
=>BC=h ⋯(2)
We know that, AB = (AC - BC)
=> 40 = (AC - BC)
=> 40=(h√3−h)[∵ from (1) & (2)]
=>40=h(√3−1)
=>h=$\dfrac{40}{(\sqrt{3}−1)}$
=$\dfrac{40}{(\sqrt{3}−1)} \times \dfrac{(\sqrt{3}+1)}{(\sqrt{3}+1)} $
=$\dfrac{40(\sqrt{3}+1)}{(3−1)}$
=$\dfrac{40(\sqrt{3}+1)}{(2)}$
=20(√3+1)
=20(1.73+1)
=20×2.73
=54.6 m
Let DC be the tower and A and B be the positions of the observer such that AB = 40 m
We have $\angle$DAC = 30°, $\angle$DBC = 45°
Let DC = h
tan30°=$\dfrac{DC}{AC}$
=>$\dfrac{1}{\sqrt{3}}=\dfrac{h}{AC}$
=>AC = h√3 ⋯(1)
tan45°=$\dfrac{DC}{BC}$
=>1=$\dfrac{h}{BC}$
=>BC=h ⋯(2)
We know that, AB = (AC - BC)
=> 40 = (AC - BC)
=> 40=(h√3−h)[∵ from (1) & (2)]
=>40=h(√3−1)
=>h=$\dfrac{40}{(\sqrt{3}−1)}$
=$\dfrac{40}{(\sqrt{3}−1)} \times \dfrac{(\sqrt{3}+1)}{(\sqrt{3}+1)} $
=$\dfrac{40(\sqrt{3}+1)}{(3−1)}$
=$\dfrac{40(\sqrt{3}+1)}{(2)}$
=20(√3+1)
=20(1.73+1)
=20×2.73
=54.6 m