question_answer

ABCD is a rectangle where the ratio of the lengths of AB and BC is 3 : 2. If P is the mid-point of AB, then the value of \[\sin \,\angle CPB\] is

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, \[\sin \,\angle CPB=\frac{BC}{PC}=\frac{2x}{\frac{5x}{2}}=\frac{4}{5}\]