Henry Ernest Dudeney/Puzzles and Curious Problems/62 - At Right Angles/Solution/Proof 1
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Puzzles and Curious Problems by Henry Ernest Dudeney: $62$
- At Right Angles
- How soon between the hours of five and six will the hour and minute hands of a clock be exactly at right angles?
Solution
- $10 \tfrac {10} {11}$ past $5$.
Proof
Let $T$ be the time in question.
Let $m$ be the number of minutes past $5:00$ that $T$ is.
Let $\theta \degrees$ be the angle made by the minute hand with respect to twelve o'clock at time $T$.
Let $\phi \degrees$ be the angle made by the hour hand with respect to twelve o'clock at time $T$.
Between $5:00$ and $6:00$ we have that $150 < \phi < 180$.
Hence from Condition for Valid Time Indication, we have that:
- $12 \paren {\phi - 150} = \theta$
Then we are told that:
- $\theta = \phi - 90$
as (from how soon) we are interested in the earliest instance of the hands being at right angles.
Hence:
\(\text {(1)}: \quad\) | \(\ds \theta + 90\) | \(=\) | \(\ds \phi\) | |||||||||||
\(\text {(2)}: \quad\) | \(\ds 12 \paren {\phi - 150}\) | \(=\) | \(\ds \theta\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 12 \paren {\theta + 90 - 150}\) | \(=\) | \(\ds \theta\) | substituting for $\phi$ in $(2)$ from $(1)$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 11 \theta\) | \(=\) | \(\ds 12 \times 60\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \theta\) | \(=\) | \(\ds \dfrac {12 \times 60} {11}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds m\) | \(=\) | \(\ds \dfrac {12 \times 60} {11} \times \dfrac 1 6\) | Speed of Minute Hand | ||||||||||
\(\ds \) | \(=\) | \(\ds 10 \tfrac {10} {11}\) | calculating |
Hence the result.
$\blacksquare$