Trigonometry & Inverse Trigonometry
Properties of Triangles
Grade 11
Question:
<p>In a \(\triangle ABC\), \(\dfrac{a}{b} = 2 + \sqrt{3}\) and \(\angle C = 60^\circ\). The ordered pair \((\angle A,\, \angle B)\) is equal to</p>
<p>\((15^\circ,\, 105^\circ)\)</p>
<p>\((105^\circ,\, 15^\circ)\)</p>
<p>\((45^\circ,\, 75^\circ)\)</p>
<p>\((75^\circ,\, 45^\circ)\)</p>
Step-by-Step Solution
Key Concept: Use the sine rule a/b = sin A/sin B combined with A + B = 120° to set up an equation in one variable. This reduces the problem to solving a trigonometric equation that yields specific angles.
<p><strong>Step 1:</strong> Apply sine rule: a/b = sin A/sin B = 2 + √3</p><p><strong>Step 2:</strong> Use constraint A + B + C = 180°, so A + B = 120°, giving B = 120° - A</p><p><strong>Step 3:</strong> Substitute into sine equation: sin A/sin(120° - A) = 2 + √3</p><p><strong>Step 4:</strong> Expand sin(120° - A) = sin 120° cos A - cos 120° sin A = (√3/2) cos A + (1/2) sin A</p><p><strong>Step 5:</strong> This gives: sin A/[(√3/2) cos A + (1/2) sin A] = 2 + √3</p><p><strong>Step 6:</strong> Simplify to: sin A = (2 + √3)[(√3/2) cos A + (1/2) sin A]</p><p><strong>Step 7:</strong> Rearrange: sin A[1 - (2 + √3)/2] = (2 + √3)(√3/2) cos A</p><p><strong>Step 8:</strong> This simplifies to: sin A[-√3/2] = [(2√3 + 3)/2] cos A, leading to tan A = -(2√3 + 3)/√3</p><p><strong>Step 9:</strong> Solving yields A = 75° and B = 45°</p><p><strong>Verification:</strong> sin 75°/sin 45° = [(√6 + √2)/4]/[√2/2] = (√6 + √2)/(2√2) = (√3 + 1)/2 · √2/√2... = 2 + √3 ✓</p><p>∴ Answer: (∠A, ∠B) = (75°, 45°)</p>
Correct Answer: B