A) 4
B) 5
C) \[11+2\sqrt{2}\]
D) 30
Correct Answer: B
Solution :
\[R\cos \theta =6\cos 0{}^\circ +2\sqrt{2}\cos ({{180}^{o}}-B)+5\cos {{270}^{o}}\] \[R\cos \theta =6-2\sqrt{2}\cos B\] ?..(i) \[R\sin \theta =6\sin 0{}^\circ +2\sqrt{2}\sin ({{180}^{o}}-B)+5\sin {{270}^{o}}\] \[R\sin \theta =2\sqrt{2}\sin B-5\] ?..(ii) From (i) and (ii), \[{{R}^{2}}=36+8{{\cos }^{2}}B-24\sqrt{2}\cos B+8{{\sin }^{2}}B\]\[+25-20\sqrt{2}\sin B\] \[=61+8({{\cos }^{2}}B+{{\sin }^{2}}B)-24\sqrt{2}\cos B-20\sqrt{2}\sin B\] \[\because \] ABC is a right angled isosceles triangle i.e., \[\angle B=\angle C=45{}^\circ \] \[\therefore \]\[{{R}^{2}}\]\[=61+8\,(1)-24\sqrt{2}\,\cdot \,\frac{1}{\sqrt{2}}-20\sqrt{2}\cdot \,\frac{1}{\sqrt{2}}\]\[=25\] \\[R=5\].You need to login to perform this action.
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