<p>The lengths of sides of a triangle are three consecutive natural numbers and its largest angle is twice the smallest one. Determine the sides of the triangle.</p>
Step-by-Step Solution
Key Concept: Use the law of sines with the constraint that the largest angle is twice the smallest angle, combined with the cosine rule to set up equations in consecutive integers.
<p><strong>Step 1:</strong> Let the sides be n, n+1, n+2 (consecutive natural numbers). Since the largest angle corresponds to the largest side, if A is the smallest angle (opposite side n) and C is the largest angle (opposite side n+2), then C = 2A.</p><p><strong>Step 2:</strong> Apply the Law of Sines: <br>sin(A)/n = sin(2A)/(n+2)<br>sin(A)/n = 2sin(A)cos(A)/(n+2)<br>Dividing by sin(A): 1/n = 2cos(A)/(n+2)<br>∴ cos(A) = (n+2)/(2n)</p><p><strong>Step 3:</strong> Apply the Cosine Rule for angle A (opposite to side n):<br>n² = (n+1)² + (n+2)² - 2(n+1)(n+2)cos(A)<br>n² = n² + 2n + 1 + n² + 4n + 4 - 2(n+1)(n+2)cos(A)<br>0 = n² + 6n + 5 - 2(n+1)(n+2)cos(A)</p><p><strong>Step 4:</strong> Substitute cos(A) = (n+2)/(2n):<br>0 = n² + 6n + 5 - 2(n+1)(n+2) · (n+2)/(2n)<br>0 = n² + 6n + 5 - (n+1)(n+2)²/n<br>Multiply by n: n³ + 6n² + 5n = (n+1)(n+2)²<br>n³ + 6n² + 5n = (n+1)(n² + 4n + 4)<br>n³ + 6n² + 5n = n³ + 4n² + 4n + n² + 4n + 4<br>n³ + 6n² + 5n = n³ + 5n² + 8n + 4<br>n² - 3n - 4 = 0<br>(n-4)(n+1) = 0</p><p><strong>Step 5:</strong> Since n must be a natural number, n = 4.<br>∴ The three sides are <strong>4, 5, 6</strong></p><p><strong>Verification:</strong> For sides 4, 5, 6: cos(A) = (25+36-16)/(2·5·6) = 45/60 = 3/4. For angle C: cos(C) = (16+25-36)/(2·4·5) = 5/40 = 1/8. Check: cos(2A) = 2cos²(A) - 1 = 2(9/16) - 1 = 1/8 ✓</p>
Correct Answer: 4, 5, 6