question_answer

A tower standing on a horizontal plane subtends   certain angle at a point 160 m apart from the foot of the tower. On advancing 100 m towards it, the tower is found to subtend an angle twice as before. The height of the tower is

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), \[\frac{1}{2}=\frac{h}{160}\]\[\Rightarrow \]\[h=\frac{160}{2}\,m\] \[\Rightarrow \]   \[h=80\,m\]