A) 80 m
B) 100 m
C) 160 m
D) 200 m
Correct Answer: A
Solution :
[a] From\[\Delta \,\Alpha \Beta C,\]\[\tan 2\theta =\frac{h}{(160-100)}\] \[\frac{2\tan \theta }{1-{{\tan }^{2}}\theta }=\frac{h}{60}\] (i) From \[\Delta \,ADC,\] \[\tan \theta =\frac{h}{160m}\] (ii) Now from Eq.(ii)\[\div \]Eq.(i), we get \[\frac{2\tan \theta }{(1-ta{{n}^{2}}\theta )\times tan\theta }=\frac{h}{60}\times \frac{160}{h}\] \[120=160\,\,(1-ta{{n}^{2}}\theta )\] \[120=160-160{{\tan }^{2}}\theta \] \[40=160{{\tan }^{2}}\theta \tan \theta =\frac{1}{2}\] Now putting the value of tan \[\theta \] in Eq, (ii), \[\frac{1}{2}=\frac{h}{160}\]\[\Rightarrow \]\[h=\frac{160}{2}\,m\] \[\Rightarrow \] \[h=80\,m\] |
You need to login to perform this action.
You will be redirected in
3 sec