question_answer

From the top of a tower, 100 m high, a man observes two cars on the opposite sides of the tower and in same straight line with its base, with angles of depression \[30{}^\circ \] and \[45{}^\circ \]. Find the distance between the cars. [Take \[\sqrt{3}=1.732\]]

Answer:

Let AB is a tower.

Cars are at point C and D respectively

In \[\Delta \,ABC\],

\[\frac{AB}{BC}=\tan \,\,30{}^\circ \]

\[\frac{100}{x}=\frac{1}{\sqrt{3}}\]

\[x=100\sqrt{3}\]

\[=100\times 1.732\]

\[=173.2\,\,m\]

In \[\Delta \,ABD\],

\[\frac{AB}{BD}=\tan \,\,45{}^\circ \]

\[\frac{100}{y}=1\]

\[y=100\,m\]

Distance between two cars \[=x+y\]

\[=173.2+100\]

\[=273.2\,m\]