A) \[20\sqrt{2}\]
B) \[10\sqrt{2}\]
C) \[10\sqrt{3}\]
D) \[20\sqrt{3}\]
Correct Answer: D
Solution :
Let the width of the road between the feet of the towers \[{{t}_{1}}\] and \[{{t}_{2}}\] be \[w\] Let the angles be \[\angle BAC=\theta \]........ [given] \[\Rightarrow \angle EBD=2\theta \] Now, from the above diagram \[\tan \theta =\frac{opposite\,\,side}{hypotenuse}=\frac{BC}{AC}\] \[\tan 2\theta =\frac{DE}{EB}\] \[BC=80-60=20\] \[AC=w\] \[DE=BO=80\], \[EB=DO=w\] \[\therefore \tan 2\theta =\frac{ED}{EB}=\frac{80}{w},\] \[\tan \theta =\frac{BC}{AC}=\frac{20}{w}\] We now that \[\tan 2\theta =\frac{2\tan \theta }{1-{{(\tan \theta )}^{2}}}\] By substituting the values of \[\tan \theta \]and \[\tan 2\theta \], we get \[\frac{80}{w}=\frac{2(\frac{20}{w})}{1-{{(\frac{20}{w})}^{2}}}\] \[\frac{40}{w}\times w=80\times [1-{{(\frac{20}{w})}^{2}}]\] \[\Rightarrow 40=80\times [1-{{(\frac{20}{w})}^{2}}]\] \[\Rightarrow [1-{{(\frac{20}{w})}^{2}}]=\frac{40}{80}=\frac{1}{2}\] \[\Rightarrow 1-\frac{400}{{{w}^{2}}}=\frac{1}{2}\] \[\Rightarrow \frac{400}{{{w}^{2}}}=\frac{1}{2}\] \[\Rightarrow 800={{w}^{2}}\] \[\Rightarrow w=\sqrt{800}=20\sqrt{2}\]
Let the width of the road the road is d. If the angle of the elevation is \[\theta \], then \[\tan \theta =\frac{20}{x},\] Here is the height difference of \[{{T}_{1}}\] and \[{{T}_{2}}\]. Given that, the angle of depression is twice of the angle of elevation. \[\tan 2\theta =\frac{60}{d}\] We know \[\tan 2\theta =\frac{2\tan \theta }{1-{{\tan }^{2}}\theta }\] \[\therefore \frac{60}{d}=\frac{40/d}{1-(400/{{d}^{2}})}\] \[\Rightarrow \frac{400}{{{d}^{2}}}=\frac{1}{3}\] \[d=20\sqrt{3}\]
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