1. **Problem statement:** We have an analogue clock with an hour hand of length 5 cm and a minute hand of length 10 cm. At 6:00, after $x$ minutes ($0 < x < 60$), the hour and minute hands form two sides of a right triangle with the third side being the segment connecting their tips. We need to find how many possible values of $x$ satisfy this. 2. **Key idea:** The hour and minute hands form two sides of lengths 5 and 10. The third side is the distance between their tips. For a right triangle, by the Pythagorean theorem, the square of one side equals the sum of the squares of the other two. 3. **Notation:** Let $\theta$ be the angle between the hour and minute hands after $x$ minutes. 4. **Lengths:** - Hour hand length $h = 5$ - Minute hand length $m = 10$ 5. **Distance between tips:** Using the law of cosines, $$d^2 = h^2 + m^2 - 2hm\cos\theta = 5^2 + 10^2 - 2 \times 5 \times 10 \cos\theta = 125 - 100\cos\theta$$ 6. **Right triangle conditions:** The right angle can be at any of the three vertices. So one of these must hold: - $h^2 + m^2 = d^2$ - $h^2 + d^2 = m^2$ - $m^2 + d^2 = h^2$ 7. **Check each case:** **Case 1:** Right angle between hour and minute hands (at the vertex connecting them): $$h^2 + m^2 = d^2$$ $$25 + 100 = d^2$$ $$125 = d^2$$ From step 5, $d^2 = 125 - 100\cos\theta$, so $$125 = 125 - 100\cos\theta \implies \cos\theta = 0$$ So $\theta = 90^\circ$ or $\pi/2$ radians. **Case 2:** Right angle at the hour hand tip: $$h^2 + d^2 = m^2$$ $$25 + d^2 = 100$$ $$d^2 = 75$$ From step 5, $$d^2 = 125 - 100\cos\theta = 75 \implies 125 - 100\cos\theta = 75 \implies \cos\theta = \frac{125 - 75}{100} = \frac{50}{100} = 0.5$$ So $\theta = 60^\circ$ or $\pi/3$ radians. **Case 3:** Right angle at the minute hand tip: $$m^2 + d^2 = h^2$$ $$100 + d^2 = 25$$ $$d^2 = -75$$ This is impossible since $d^2$ cannot be negative. So no solution here. 8. **Summary of angle conditions:** - $\cos\theta = 0$ (i.e., $\theta = 90^\circ$) - $\cos\theta = 0.5$ (i.e., $\theta = 60^\circ$) 9. **Find $\theta$ as a function of $x$:** - The minute hand moves $6^\circ$ per minute. - The hour hand moves $0.5^\circ$ per minute. At 6:00, the hour hand is at $180^\circ$ (pointing down). After $x$ minutes: - Hour hand angle: $180 + 0.5x$ - Minute hand angle: $6x$ The angle between them is $$\theta = |(180 + 0.5x) - 6x| = |180 - 5.5x|$$ Since angles wrap every $360^\circ$, consider $\theta$ modulo $360^\circ$ and take the smaller angle between hands: $$\theta = \min(|180 - 5.5x|, 360 - |180 - 5.5x|)$$ 10. **Solve for $x$ when $\theta = 90^\circ$:** $$|180 - 5.5x| = 90$$ Two cases: - $180 - 5.5x = 90 \implies 5.5x = 90 \implies x = \frac{90}{5.5} = \frac{180}{11} \approx 16.36$ - $180 - 5.5x = -90 \implies 5.5x = 270 \implies x = \frac{270}{5.5} = \frac{540}{11} \approx 49.09$ Both are in $(0,60)$. 11. **Solve for $x$ when $\theta = 60^\circ$:** $$|180 - 5.5x| = 60$$ Two cases: - $180 - 5.5x = 60 \implies 5.5x = 120 \implies x = \frac{120}{5.5} = \frac{240}{11} \approx 21.82$ - $180 - 5.5x = -60 \implies 5.5x = 240 \implies x = \frac{240}{5.5} = \frac{480}{11} \approx 43.64$ Both are in $(0,60)$. 12. **Count total solutions:** - For $\theta=90^\circ$: 2 values - For $\theta=60^\circ$: 2 values Total $= 4$ possible values of $x$. **Final answer:** (D) 4