Question

A girl travels along a straight line, from point A to B at a constant speed, $V_1$ meters/sec for T seconds. Next, she travels from point B to C along a straight line, at a constant speed of $V_2$ meters/sec for another T seconds. BC makes an angle 105° with AB. If CA makes an angle 30° with BC, how much time will she take to travel back from point C to A at a constant speed of $V_2$ meters/sec, if she travels along a straight line from C to A?

Accepted Answer

C. In $	riangle ABC$: $AB = V_1 T$, $BC = V_2 T$. Interior angle $\angle B = 180^\circ - 105^\circ = 75^\circ$. $\angle C = 30^\circ$. Sum of angles = $180^\circ \implies \angle A = 180^\circ - 75^\circ - 30^\circ = 75^\circ$. Since $\angle A = \angle B = 75^\circ$, $	riangle ABC$ is isosceles with $AC = BC = V_2 T$ or using Sine Rule: $\frac{AC}{\sin 75^\circ} = \frac{BC}{\sin 75^\circ} \implies AC = BC = V_2 T$. Wait, using altitude $BD \perp AC$: $BD = V_2 T \sin 30^\circ = \frac{V_2 T}{2}$. $CD = V_2 T \cos 30^\circ = \frac{\sqrt{3}}{2} V_2 T$. In $	riangle ABD$, $\angle A = 45^\circ \implies AD = BD = \frac{V_2 T}{2}$. Total $AC = CD + AD = \frac{\sqrt{3} + 1}{2} V_2 T = 0.5(\sqrt{3} + 1) V_2 T$. Time taken at speed $V_2 = \frac{AC}{V_2} = 0.5(\sqrt{3} + 1)T$.

Triangular Geometry and Sine Rule

Explanation

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