(a) \( \angle JDL = 99^\circ \) (M1A1, N2).

Working:
Attempt to use bearing and angle relationships (M1)
Bearing of D from J is \( 034^\circ \), so \( \angle NJD = 34^\circ \) (N is North).
\( \angle NJL = 90^\circ \) (Jaipur to Lucknow is East).
\( \angle DJL = 90^\circ – 34^\circ = 56^\circ \).
Given \( \angle JLD = 25^\circ \).
Sum of angles in triangle \( JDL \): \( \angle JDL = 180^\circ – 56^\circ – 25^\circ = 99^\circ \) (A1).

[2 marks]

(b) \( DL \approx 420 \, \text{km} \) (M1A1A1, N3).

Working:
Attempt to apply the sine rule (M1)
In triangle \( JDL \): \( \frac{DL}{\sin \angle DJL} = \frac{JL}{\sin \angle JDL} \).
Given \( JL = 500 \, \text{km} \), \( \angle DJL = 56^\circ \), \( \angle JDL = 99^\circ \).
Substitute: \( \frac{DL}{\sin 56^\circ} = \frac{500}{\sin 99^\circ} \).
Solve: \( DL = \frac{500 \times \sin 56^\circ}{\sin 99^\circ} \).
Approximate: \( \sin 56^\circ \approx 0.8290 \), \( \sin 99^\circ \approx 0.9877 \).
Calculate: \( DL \approx \frac{500 \times 0.8290}{0.9877} \approx 420 \, \text{km} \) (A1A1).

[3 marks]

Total [5 marks]