Question

Two men are on opposite sides of a tower. tthey measure the angle of elevation of the top of the tower is 30° and 45° is the height of the tower is 50 m, find the distance between the two men .

Answer 1

$\text{Let A and B be the positions of two men}$

$\text{Let XY be the tower}$

$\textbf{Given:}$

$XY=50\;m$

$\text{Angle of elevations 30^{\circ} and 45^{\circ} }$

$\textbf{To find:}$

$\text{Distance between the two men}$

$\textbf{Solution:}$

$\text{In \triangle AYX,}$

$\cot\,45^{\circ}=\dfrac{AY}{XY}$

$1=\dfrac{AY}{50}$

$\implies\bf\,AY=50$

$\text{In \triangle BYX,}$

$\cot\,30^{\circ}=\dfrac{BY}{XY}$

$\sqrt{3}=\dfrac{BY}{50}$

$\implies\,BY=50{\times}\sqrt{3}$

$\implies\,BY=50{\times}1.732$

$\implies\bf\,BY=86.60$

$\text{Now,}$

$=AB$

$=AY+BY$

$=50+86.60$

$=136.60\;m$

$\therefore\textbf{Distance between the two men is 136.60 meters}$

Answer 2

Answer:

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