XAT 2015 — QA & DI Question 19

Let m∠DAB = θ ⇒ m∠BCD = 2θ
□PBCD is a parallelogram.
∴ m∠DPB = 2θ
m∠PBC = m∠PDC = (180 – 2θ)
∠DPB is an exterior angle of ∆PAB.
∴ By exterior angle theorem, m∠PBA = θ
as, m∠DAB = θ
Thus, in ∆PAB, PA = PB

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∴ m∠ABC = θ + (180 – 2θ) = 180 – θ
According to the given conditions, 10x + y = 1120 and 10x = 1000
Solving the two equations, we get x = 100 and y = 120
sin (180 – θ) = sin θ
Applying sine rule to ∆PAB,

$\frac{100}{\sin \mathrm{\theta }}=\frac{120}{\sin (180-2\mathrm{\theta })}$

$$\begin{gathered} \frac{}{} \\ \frac{100}{\sin \mathrm{\theta }}=\frac{120}{\sin 2\mathrm{\theta }} .....\left(\mathrm{i}\right) \end{gathered}$$

sin 2θ = 2 × sin θ × cos θ

∴ (i) becomes

$\frac{100}{\sin \theta }=\frac{120}{2\times \sin \theta \times \cos \theta }$

$$\begin{gathered} \frac{}{} \\ \therefore \cos \theta =\frac{3}{5}\Rightarrow \sin \theta =\frac{4}{5} \end{gathered}$$

$$\begin{gathered} \frac{\mathrm{}}{} \\ \therefore \sin \angle ABC=\sin (180-\theta )=\sin \theta =\frac{4}{5} \end{gathered}$$

Hence, option (a).