Question

The parallel sides of a trapezoid ABCD are in the ratio of $4 : 5$. ABCD is divided into an isosceles triangle ABP and a parallelogram PBCD. ABCD has a perimeter equal to $1120	ext{ meters}$ and PBCD has a perimeter equal to $1000	ext{ meters}$. Find $\sin \angle ABC$, given $2\angle DAB = \angle BCD$.

Accepted Answer

A. Using perimeter relations: $AP = AB = 120$, $BP = 100$. In $	riangle ABP$, applying cosine rule for $\angle A$ gives $\cos A = 3/5 \implies \sin A = 4/5$. Since $\angle ABC = 180^\circ - \angle A$, $\sin \angle ABC = \sin A = 4/5$.

Trapezoid Side Ratios and Sine Function

Explanation

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