Vanuatu
From the top of a tower ^@ h \space meter ^@ high, the angle of depression of two objects, which are in line to the foot of the tower is ^@ \alpha ^@ and ^@ \beta ^@ ^@( \beta > \alpha)^@. Find the distance between the two objects.
Answer:
^@(\cot \alpha - \cot \beta) h \space meters ^@
- Let ^@ AB ^@ be the tower of height ^@ h \space meter ^@ and ^@ x \space meter ^@ be the distance between the two objects ^@ C ^@ and ^@ D ^@.
As ^@ \beta > \alpha ^@, ^@ \beta ^@ will be the angle of depression of the point ^@ D ^@ and ^@ \alpha ^@ will be the angle of depression of the point ^@ C. ^@
The situation given in the question is represented by the image given below.
- In the right-angled triangle ^@ ABD ^@, we have @^ \begin{aligned} & \tan \beta = \dfrac { AB } { AD } \\ \implies & \tan \beta = \dfrac { h } { AD } \\ \implies & AD = \dfrac { h } { \tan \beta } \\ \implies & AD = h \space cot \beta && \bigg[\text{As,} \cot \beta = \dfrac { 1 } { \tan \beta } \bigg] && \ldots \text{(i)} \\ \end{aligned} @^
- In right-angled triangle ^@ ABC ^@, we have @^ \begin{aligned} & \tan \alpha = \dfrac { AB } { AC } \\ \implies & \tan \alpha = \dfrac { h } { AC } \\ \implies & AC = \dfrac { h } { \tan \alpha } \\ \implies & AD + x = h \space \cot \alpha && \bigg[ \text{ As,} \cot \alpha = \dfrac { 1 } { \tan \alpha } \text{ and AC = AD + x } \bigg] && \ldots \text{(ii)} \\ \end{aligned} @^
- Now, let us subtract ^@ eq \space \text{(i)} ^@ from ^@ eq \space \text{(ii)} ^@. @^ \begin{aligned} & (AD + x) - AD = h \space cot \alpha - h \space cot \beta \\ \implies & x = (cot \alpha - cot \beta) h \\ \end{aligned} @^
- Therefore, the distance between two objects is ^@ (\cot \alpha - \cot \beta) h \space meters ^@.
