First slide

Applications of trigonometry

Question

A vertical pole (more than 50 meters high) consists of two portions. The lower being $\frac{1}{3} r d$ of the whole. If the upper portion subtends an angle $\tan^{- 1} (\frac{1}{2})$ at a point in a horizontal plane through the foot of the pole at a distance 40 meters from it, then the height of the pole is

Moderate

A

80m
B

90m
C

70m
D

120m

Solution

Let AB = h meter, BC = 2h meter
$∠ A P C = α, ∠ A P B = β, ∠ B P C = θ$ so that $θ = α - β$
It is given that $\tan θ = \frac{1}{2}$
and $A P = 40 m$ , where we have $\tan α = \frac{3 h}{40}, \tan β = \frac{h}{40}$
$N o w \tan θ = \tan (α - β) = \frac{\tan α - \tan β}{1 + \tan α + \tan β}$
$\Rightarrow \frac{1}{2} = \frac{\frac{3 h}{40} - \frac{h}{40}}{1 + (\frac{3 h}{40}) (\frac{h}{40})} = \frac{80 h}{1600 + 3 h^{2}}$
$\Rightarrow 3 h^{2} - 160 h + 1600 = 0 \Rightarrow (3 h - 40) (h - 40) = 0$
$\Rightarrow h = 40$ , and hence $3 h = 120$ is the required length of the pole. Note that as the pole is given to $3 h \neq 40$ be more than 50 meters height.

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