question_answer

ABC is an isosceles triangle right angled at A. Forces of magnitude \[2\sqrt{2,}\,5\] and 6 act along \[\overrightarrow{BC},\,\,\overrightarrow{CA}\] and \[\overrightarrow{AB}\] respectively. The magnitude of their resultant force is                                                   [Roorkee 1999]

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\].