question_answer

    A tower subtends an angle a at a point A in the plane of its base and the angle of depression of the foot of the tower at a point b ft just above A is P. Then, height of the tower is

A)  \[b\text{ }tan\text{ }\alpha \text{ }cot\text{ }\beta \]                  

B)  \[b~cot\text{ }\alpha \text{ }tan\text{ }\beta \]

C)  \[b\text{ }tan\text{ }\alpha \text{ }tan\text{ }\beta \]                  

D)  \[b\text{ }cot\text{ }\alpha \text{ }cot\text{ }\beta \]

Correct Answer: A

Solution :

                Let the height of tower PQ is h and let point A is at a distance x from the foot\[p\]of the tower PQ. Now, ln \[\Delta APQ\]                 \[\tan \alpha =\frac{h}{x}\] \[\Rightarrow \]               \[h=x\tan \alpha \]                                    ...(i) In \[\Delta ABP,\]                 \[\tan \beta =\frac{b}{x}\] \[\Rightarrow \]               \[b=x\tan \beta \] \[\Rightarrow \]               \[x=\frac{b}{\tan \beta }\] Putting this value of\[x\]is Eq. (i), we get \[h=\frac{b}{\tan \beta }.\tan \alpha \] \[\Rightarrow \]      \[h=b\tan \alpha \cot \beta \]