A) \[\frac{h\,\,\tan \,\,\alpha }{\tan \,\,\beta +\tan \,\,\alpha }\]
B) \[\frac{h\,\,\tan \,\,\alpha }{\tan \,\,\beta -\tan \,\,\alpha }\]
C) \[\frac{h\,\,\tan \,\,\alpha }{\tan \,\,\beta }\]
D) \[\frac{h\,\,\tan \,\,\beta }{\tan \,\,\beta -\tan \,\,\alpha }\]
Correct Answer: B
Solution :
(b): In \[\Delta \,ACD,\,\,tan\beta =\frac{x+h}{y}\] \[y=\frac{x+h}{\tan \beta }\] ?..(1) In \[\Delta BCD,\tan \alpha =\frac{x}{y}\] \[\therefore y=\frac{x}{\tan \alpha }\] ?..(2) From (1) and (2), \[\frac{x+h}{\tan \beta }=\frac{x}{\tan \alpha }\] \[\Rightarrow \] \[x\,\tan \alpha +h\tan \alpha =x\tan \beta \] \[\Rightarrow \] \[h\tan \alpha =x(tan\beta -tan\alpha )\] \[\Rightarrow \] \[x=\frac{h\,\,\tan \alpha }{\tan \beta -\tan \alpha }\]You need to login to perform this action.
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