At what time between $4$ o'clock and $5$ o'clock will the hands of an analog clock be at right angles ($90^\circ$) to each other for the FIRST time? Express the time as $4\text{ hours and } M\text{ minutes}$. What is the value of $M$? ________ minutes (round off to two decimal places).
Answer Key:
5.45
Step-by-Step Solution & Explanation
**Method 1: Relative Angular Velocity Formulation**
At 4:00, the hour hand is at $30^\circ \times 4 = 120^\circ$ and the minute hand is at $0^\circ$.
For the hands to be at $90^\circ$ for the first time, the minute hand must reduce the initial separation of $120^\circ$ to $90^\circ$, gaining $120^\circ - 90^\circ = 30^\circ$ on the hour hand.
Relative speed of the minute hand $= 5.5^\circ/\text{min} = \frac{11}{2}^\circ/\text{min}$.
$M = \frac{30^\circ}{11/2^\circ/\text{min}} = \frac{60}{11} = 5.4545...\text{ minutes} \approx 5.45\text{ minutes}$
**Method 2: Formula Verification**
$\theta = |30H - 5.5M| = 90 \implies 120 - 5.5M = 90 \implies 5.5M = 30 \implies M = \frac{60}{11} \approx 5.45$