Hong Kong
From a window ^@h \space meters ^@ high above the ground in a street, the angles of elevation and depression of the top and foot of the other house on the opposite side of the street are ^@\alpha^@ and ^@\beta^@ respectively. Show that the height of the opposite house is ^@ h (1 + \tan \alpha \space cot \beta) \space meters^@.
Answer:
- Let ^@W^@ be the window and ^@AB^@ be the house on the opposite side with height ^@(h + h') \space meters^@.
The figure below shows the given situation. - In the right-angled triangle ^@AWP^@, we have @^ \begin{aligned} & \cot \alpha = \dfrac { WP } { AP } \\ \implies & \cot \alpha = \dfrac { WP } { h' } \\ \implies & WP = h' \cot \alpha && \ldots \text{(i)} \end{aligned} @^
- In right-angled triangle ^@WPB^@, we have @^ \begin{aligned} & \cot \beta = \dfrac { WP } { BP } \\ \implies & \cot \beta = \dfrac { WP } { h } \\ \implies & WP = h \cot \beta && \ldots \text{(ii)} \end{aligned} @^
- On comparing ^@eq \space \text{(i)} ^@ and ^@eq \space \text{(ii)}^@, we get @^ \begin{aligned} & h' \cot \alpha = h \cot \beta \\ \implies & h' = \dfrac { h \cot \beta } { \cot \alpha } = h \tan \alpha \cot \beta \end{aligned} @^
- Thus, height of the house = ^@ h + h' = h + h \tan \alpha \cot \beta = h (1 + \tan \alpha \cot \beta) \space meters^@.
