A) \[\frac{u}{g}s\]
B) \[\frac{2u}{g}s\]
C) \[\frac{u}{2g}s\]
D) None of these
Correct Answer: B
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 \[h=\frac{b}{\tan \beta }.\tan \alpha \] \[\Rightarrow \] \[h=b\tan \alpha \cot \beta \]You need to login to perform this action.
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