Question

The angle of elevation of the top of a tower as observed from a point in a horizontal plane through the foot of the tower is 32°.When the observer moves towards the tower a distance of 100m,he finds the angle of elevation of the top to be 63°.Find the height of the tower and the distance of the first position from the tower.(tan32°=0.6248 and tan 63°=1.9626)

Answer 1

$\text{Let AB be the tower}$

$\text{Let P and Q are first and second point of observations}$

$\textbf{Given:}$

$PQ=100\;m$

$\textbf{To find:}$

$\text{Height of the tower and distance of the first position from the tower}$

$\textbf{Solution:}$

$\text{In \triangle ABQ,}$

$\tan{63^{\circ}}=\dfrac{AB}{BQ}$

$1.9626=\dfrac{h}{x}$

$\implies\bf\,h=1.9626\,x$ ....(1)

$\text{In \triangle ABP,}$

$\tan{32^{\circ}}=\dfrac{AB}{BP}$

$0.6248=\dfrac{h}{x+100}$

$\implies\bf\,]h=0.6248\,x+62.48$ .......(2)

$\text{Using (1) in (2), we get}$

$1.9626\,x=0.6248\,x+62.48$

$1.9626\,x-0.6248\,x=62.48$

$1.3378\,x=62.48$

$\implies\,x=\dfrac{62.48}{1.3378}$

$\implies\boxed{\bf\,x=46.7}$

$\text{From (1),}$

$h=1.9626{\times}46.7$

$h=91.67$

$\text{Also,}$

$BP=BQ+QP$

$BP=x+100$

$BP=46.7+100$

$\implies\boxed{\bf\,BP=146.7}$

$\textbf{Answer:}$

$\textbf{Height of the tower is 91.67 m}$

$\textbf{Distance of the first position from the tower is 146.7 m}$

Answer 2

Step-by-step explanation:

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