question_answer

P is a point on the segment joining the feet of two vertical poles of heights\[a\]and\[b\]. The angles of elevation of the tops of the poles from P are\[{{45}^{o}}\]each. Then, the square of the distance between the tops of the poles is

A) \[\frac{{{a}^{2}}+{{b}^{2}}}{2}\]               

B) \[{{a}^{2}}+{{b}^{2}}\]

C) \[2({{a}^{2}}+{{b}^{2}})\]            

D)        \[4({{a}^{2}}+{{b}^{2}})\]

Correct Answer: C

Solution :

In\[\Delta \,\,APD\], \[\tan {{45}^{o}}=\frac{a}{AP}\Rightarrow AP=a\] and in\[\Delta \,\,BPC\],                 \[\tan {{45}^{o}}=\frac{b}{PB}\]  \[\Rightarrow \]   \[PB=b\] \[\therefore \]  \[DE=a+b\]   and   \[CE=b-a\] In\[\Delta \,\,DEC\],                 \[D{{C}^{2}}=D{{E}^{2}}+E{{C}^{2}}\]                 \[={{(a+b)}^{2}}+{{(b-a)}^{2}}\]                 \[=2({{a}^{2}}+{{b}^{2}})\]