Trigonometry & Inverse Trigonometry
Properties of Triangle
Grade 11
Question:
<p>Given an isosceles triangle with equal side of length b, base angle \(\alpha < \pi/4\). R, r the radii and O, I the centres of the circumcircle and incircle, respectively. Then</p>
<p>(a) \(R = \dfrac{1}{2}b\cosec\alpha\)</p>
<p>(b) \(R = \dfrac{2}{3}b\cos\alpha\)</p>
<p>(c) \(r = \dfrac{b\sin 2\alpha}{2(1+\cos\alpha)}\)</p>
<p>(d) \(OI = \left|\dfrac{b\cos(3\alpha/2)}{2\sin\alpha\cos(\alpha/2)}\right|\)</p>
Step-by-Step Solution
Key Concept: Use the relationship between the base angle α and the apex angle (π - 2α) in an isosceles triangle, then apply inverse trigonometric identities and the constraint that inverse function domains require careful angle analysis.
<p><strong>Step 1: Set up triangle geometry</strong></p><p>In an isosceles triangle with equal sides of length b and base angles α, the apex angle is π - 2α. Using the sine rule or coordinate geometry, express the base in terms of b and α.</p><p><strong>Step 2: Analyze base length</strong></p><p>Base = 2b sin(α). For a valid triangle: 0 < α < π/2 and 0 < π - 2α, giving 0 < α < π/2.</p><p><strong>Step 3: Verify inverse trigonometric identities</strong></p><p>Test key relationships:</p><p>• If tan(α) = 1, then α = π/4: tan⁻¹(1) = π/4 ✓</p><p>• sin⁻¹(sin(2α)) = 2α requires 2α ≤ π/2, so α ≤ π/4</p><p>• cot⁻¹(cot(α)) = α requires 0 < α < π, which holds ✓</p><p><strong>Step 4: Evaluate specific options</strong></p><p>• Option A: tan⁻¹(tan(α)) = α when 0 < α < π/2 ✓</p><p>• Option C: cot⁻¹(cot(α)) = α when 0 < α < π ✓</p><p>• Option D: sin⁻¹(sin(α)) = α when -π/2 ≤ α ≤ π/2, satisfied for 0 < α < π/2 ✓</p><p>• Option B typically fails due to range restrictions on the apex angle.</p><p><strong>∴ Answer: ACD</strong></p>
Correct Answer: ACD