From a point on the ground, the angle of elevation of the top of a vertical tower is found to be such that its tangent is 35. On walking 50 m towards the tower, the tangent of the new angle of elevation of the top of the tower is found to be 45. Find the height of the tower.

Let AB is the tower and AB=hP, Q are the observation pointsGiven PQ=50 mtanα=35, tanβ=45In ΔABPIn ΔABQtanα=ABPBtanβ=ABBQ35=h50+x45=hx5h=150+3x→(1)5h=4x→(2)From (1) and (2)4x=150+3xx=150 mSub x=150 in (1)5h=150+4505h=600h=120 m

From a point on the ground, the angle of elevation of the top of a vertical tower is found to be such that its tangent is 35. On walking 50 m towards the tower, the tangent of the new angle of elevation of the top of the tower is found to be 45. Find the height of the tower.

Let AB is the tower and AB=hP, Q are the observation pointsGiven PQ=50 mtanα=35, tanβ=45In ΔABPIn ΔABQtanα=ABPBtanβ=ABBQ35=h50+x45=hx5h=150+3x→(1)5h=4x→(2)From (1) and (2)4x=150+3xx=150 mSub x=150 in (1)5h=150+4505h=600h=120 m

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From a point on the ground, the angle of elevation of the top of a vertical tower is found to be such that its tangent is $\frac{3}{5}$ . On walking $50 m$ towards the tower, the tangent of the new angle of elevation of the top of the tower is found to be $\frac{4}{5}$ . Find the height of the tower.

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$121 m$

$150 m$

$120 m$

$160 m$

answer is D.

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Detailed Solution

Let $A B$ is the tower and $A B = h$

$P, Q$ are the observation points

Given $P Q = 50 m$

$\tan α = \frac{3}{5}, \tan β = \frac{4}{5}$

$\begin{matrix} In Δ A B P & In Δ A B Q \\ \tan α = \frac{A B}{P B} & \tan β = \frac{A B}{B Q} \\ \frac{3}{5} = \frac{h}{50 + x} & \frac{4}{5} = \frac{h}{x} \\ 5 h = 150 + 3 x \to (1) & 5 h = 4 x \to (2) \end{matrix}$