At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is \frac{5}{12}. On walking 192 {m} towards the tower, the tangent of the angle of elevation is \frac{3}{4}. The height of the tower is

Management Aptitude Test Sep 2020 Paper

Question : 102 of 200

Marks: +1, -0

At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is

\;{5}/{12}

. On walking

192 {m}

towards the tower, the tangent of the angle of elevation is

\;{3}/{4}

. The height of the tower is

Solution:

Let the height of tower be

h {m}

According to the question,

\;\; \tan θ_{1}=\;{h}/{192+B C}

⇒ \;\; \;{5}/{12}=\;{h}/{192+B C}

. . . (i)

⇒ \;\; \;\; \;{12}/{5}=\;{192+B C}/{h}

\;\text " and "\; \;\; \tan θ_{2}=\;{h}/{B C}

⇒ \;\; \;{B C}/{h}=\;{4}/{3}

. . . (ii)
On subtracting Eq. (ii) from Eq. (i), we get

\;{192}/{h} \;\; =\;{12}/{5}-\;{4}/{3}=\;{36-20}/{15}=\;{16}/{15}

∴ \;\; \;\; h \;\; =\;{192 × 15}/{16}=180 {m}

Go to Question:

Prev Question

Next Question