A) \[{{30}^{{}^\circ }}\]
B) \[{{45}^{{}^\circ }}\]
C) \[{{60}^{{}^\circ }}\]
D) None of these
Correct Answer: C
Solution :
\[PA=PR\] \[\angle APB=\angle ARB={{45}^{o}}\] If \[PR=\sqrt{2x}\,:\] \[PB=\frac{x}{\sqrt{2}}\] From \[\Delta \,APB,\] \[\tan {{45}^{o}}=\frac{AB}{PB}\Rightarrow AB=PB=\frac{x}{\sqrt{2}}\] \[\therefore \] \[PA=\sqrt{\frac{{{x}^{2}}}{2}+\frac{{{x}^{2}}}{2}}=\sqrt{{{x}^{2}}}=x\] \[\therefore \] \[QA=PQ=PA=x\] \[\therefore \] \[\angle PAQ={{60}^{o}}\]You need to login to perform this action.
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