A) 100 ft
B) 120 ft
C) 150 ft
D) none of these
Correct Answer: B
Solution :
Let the height of the pole BC be \[h.\] In \[\Delta \,ABC,\] \[\tan \beta =\frac{h}{40}\] ?(i) and in\[\Delta \Alpha \Beta D,\] \[\tan \alpha =\frac{h/3}{40}\] \[=\frac{h}{120}\] ?(ii) Now, \[\tan \theta =\frac{1}{2}(given)\] \[\Rightarrow \] \[\tan (\beta -\alpha )=\frac{1}{2}\] \[\Rightarrow \] \[\frac{\tan \beta -\tan \alpha }{1+\tan \beta \tan \alpha }=\frac{1}{2}\] \[\Rightarrow \] \[\frac{\frac{3h}{120}-\frac{h}{120}}{1+\frac{3{{h}^{2}}}{14400}}=\frac{1}{2}\]\[\Rightarrow \]\[h=120,40\] But \[h\]cannot be taken according to the given condition, therefore \[h=120\,ft.\]You need to login to perform this action.
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