question_answer

Two points A and B are on the same side of a tower and in the same straight line with its base. The angles of depression of these points from the top of the tower are \[60{}^\circ \] and \[45{}^\circ \] respectively. If the height of the tower is 15 m, then find the distance between these points.

Answer:

Let PT be tower

From \[\Delta \text{ }PTA\]

\[\tan \,\,60{}^\circ =\frac{PT}{TA}\Rightarrow TA=\frac{15}{\sqrt{3}}\]

from \[\Delta \text{ }PTB\]

\[\tan \,\,45{}^\circ =\frac{PT}{TB}\Rightarrow TB=PT=15\,\,m\]

Distance between two points

\[AB=TB-TA\]

\[=15-\frac{15}{\sqrt{3}}=\frac{15\left( \sqrt{3}-1 \right)}{\sqrt{3}}m\]