A) \[\frac{3}{5}\]
B) \[\frac{2}{5}\]
C) \[\frac{3}{4}\]
D) \[\frac{4}{5}\]
Correct Answer: D
Solution :
[d] Let \[AB=3x,\]\[BC=2x.\] After drawing \[PQ\parallel BC,\] we get the rectangle PBCQ where \[PB=\frac{3x}{2},\]\[BC=2x\] \[\therefore \]\[PC=\sqrt{P{{B}^{2}}+B{{C}^{2}}}\] \[\sqrt{\frac{9{{x}^{2}}}{4}}+4{{x}^{2}}=\sqrt{\frac{25{{x}^{2}}}{4}}=\frac{5x}{2}\] Now, \[\sin \,\angle CPB=\frac{BC}{PC}=\frac{2x}{\frac{5x}{2}}=\frac{4}{5}\] |
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