Rectangle with midpoint and triangle at P — evaluate sin(∠CPB): In rectangle ABCD, AB : BC = 3 : 2. Point P is the midpoint of AB. What is the value of sin(∠CPB)?

Introduction / Context:
Coordinate geometry streamlines trig in plane figures. Placing the rectangle on axes lets us compute vectors for the sides that define ∠CPB and then evaluate sin using the cross-product formula for the sine of the angle between two vectors.

Given Data / Assumptions:

• AB : BC = 3 : 2; take convenient lengths AB = 3 and BC = 2 (units).
• Let A = (0, 0), B = (3, 0), C = (3, 2), D = (0, 2).
• P is the midpoint of AB ⇒ P = (1.5, 0).

Concept / Approach:
At P, form vectors to B and C: PB and PC. For vectors u and v, sin θ = |u × v| / (|u||v|), where in 2D the cross magnitude is |x_uy_v − y_ux_v|.

Step-by-Step Solution:
PB = B − P = (3 − 1.5, 0 − 0) = (1.5, 0)PC = C − P = (3 − 1.5, 2 − 0) = (1.5, 2)|PB| = √(1.5² + 0²) = 1.5; |PC| = √(1.5² + 2²) = √6.25 = 2.5|PB × PC| = |1.5·2 − 0·1.5| = 3sin(∠CPB) = 3 / (1.5 × 2.5) = 3 / 3.75 = 0.8 = 4/5

Verification / Alternative check:
Using dot/cross formulas yields the same 0.8. Ratio choice does not affect result because any positive scalar multiple keeps the angle unchanged.

Why Other Options Are Wrong:
3/5 (0.6), 2/5 (0.4), 3/4 (0.75), and 1/2 (0.5) do not match the computed 0.8.

Common Pitfalls:
Using length ratio AB/BC directly inside trigonometric functions or mixing up the vectors’ order (which flips the sign but not the magnitude for sine).

Final Answer:
4 5