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Explanation:

At 11 am, the car is at C and Ayesha is at A.
In △ABC, angle ACB = 45°
Tan45° = 1 = $\frac{\mathrm{AB}}{\mathrm{BC}}$
⇒ BC = AB = 200

At 11:02 am, the car is at D and Ayesha is at A.
In △ABD, angle ADB = 30°
Tan30° = $\frac{1}{\sqrt{3}}$ = $\frac{\mathrm{AB}}{\mathrm{BD}}$
⇒ BD = AB√3 = 200√3

⇒ CD = 200√3 - 200 = 200(√3 - 1) = 200(0.732) = 146.4 m

∴ The car moved 146.4 m in 2 minutes.

⇒ Speed of the car = $\frac{0.1464 \mathrm{kms}}{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$60$}\right.} \mathrm{hours}}$ = 4.392 kmph.

Hence, option (d).