ABCD is a rectangle where the ratio of the le | vedclass

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(D) Let $AB = 3x$ units and $BC = 2x$ units.Since $P$ is the midpoint of $AB$,$PB = \frac{AB}{2} = \frac{3x}{2}$ units.In the right-angled triangle $\

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$\frac{4}{5}$

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(D) Let $AB = 3x$ units and $BC = 2x$ units.Since $P$ is the midpoint of $AB$,$PB = \frac{AB}{2} = \frac{3x}{2}$ units.In the right-angled triangle $	riangle CPB$,by the Pythagorean theorem:$CP = \sqrt{PB^2 + BC^2} = \sqrt{(\frac{3x}{2})^2 + (2x)^2}$$CP = \sqrt{\frac{9x^2}{4} + 4x^2} = \sqrt{\frac{9x^2 + 16x^2}{4}} = \sqrt{\frac{25x^2}{4}} = \frac{5x}{2}$ units.Now,in $	riangle CPB$,$\sin(\angle CPB) = \frac{	ext{Opposite side}}{	ext{Hypotenuse}} = \frac{BC}{CP}$.$\sin(\angle CPB) = \frac{2x}{\frac{5x}{2}} = 2x \cdot \frac{2}{5x} = \frac{4}{5}$.

$ABCD$ is a rectangle where the ratio of the lengths of $AB$ and $BC$ is $3:2$. If $P$ is the midpoint of $AB$,then the value of $\sin(\angle CPB)$ is

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