A) 13, 5
B) 12, 6
C) 14, 4
D) 11, 7
Correct Answer: A
Solution :
\[P+Q=18,\,\,R=12,\,\,\theta ={{90}^{o}}\], (say) \[\tan \,\theta =\tan {{90}^{o}}=\infty \] \[\Rightarrow \,\,\,P+Q\,\cos \,\alpha =0\], \[\therefore \,\,\cos \alpha =\frac{-P}{Q}\] Also, \[{{(12)}^{2}}={{P}^{2}}+{{Q}^{2}}+2PQ\,\,\cos \alpha \] or \[144={{P}^{2}}+{{Q}^{2}}+(2P)(-P)\] \[\Rightarrow \,\,\,144={{Q}^{2}}-{{P}^{2}}=(Q+P)(Q-P)\] or \[144=18\,(Q-P)\] or \[Q-P=8\] After solving Q = 13, P = 5.You need to login to perform this action.
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