Trigonometry & Inverse Trigonometry
Properties of Triangle
Grade None
Question:
<p>If the sines of the angles A and B of a triangle ABC satisfy the equation \(c^2x^2 - c(a+b)x + ab = 0\), then the triangle</p>
<p>(a) is acute angled</p>
<p>(b) is right angled</p>
<p>(c) is obtuse angled</p>
<p>(d) satisfy \(\sin A + \cos A = \dfrac{(a+b)}{c}\)</p>
Step-by-Step Solution
Key Concept: Since sin A and sin B are roots of the given quadratic equation, use Vieta's formulas to relate them to the sides via the sine rule (a/sin A = b/sin B = c/sin C), then apply the constraint that A + B + C = π.
<p><strong>Step 1:</strong> Since sin A and sin B satisfy the equation c²x² - c(a+b)x + ab = 0, by Vieta's formulas:</p><p>sin A + sin B = (a+b)/c</p><p>sin A · sin B = ab/c²</p><p><strong>Step 2:</strong> Apply the sine rule: a/sin A = b/sin B = c/sin C = 2R (where R is circumradius)</p><p>This gives: sin A = a/(2R), sin B = b/(2R), sin C = c/(2R)</p><p><strong>Step 3:</strong> Substitute into Vieta's relations:</p><p>a/(2R) + b/(2R) = (a+b)/c ⟹ (a+b)/(2R) = (a+b)/c ⟹ c = 2R</p><p><strong>Step 4:</strong> From sine rule: sin C = c/(2R) = 2R/(2R) = 1</p><p>Therefore: C = π/2</p><p><strong>Step 5:</strong> Verify the second relation: sin A · sin B = ab/(4R²) = ab/c² ✓ (This confirms consistency)</p><p>∴ The triangle is <strong>right-angled at C</strong> (or isosceles right-angled if additional constraints apply)</p><p><strong>Answer: BD</strong> (right-angled triangle)</p>
Correct Answer: BD