Trigonometry & Inverse Trigonometry
Special triangles
Grade 11
Question:
<p>Circumradius of an isosceles △ABC with ∠A = ∠B is 4 times its inradius, then cos A is root of the equation:</p>
<p>(a) \(x^2 - x - 8 = 0\)</p>
<p>(b) \(8x^2 - 8x + 1 = 0\)</p>
<p>(c) \(x^2 - x - 4 = 0\)</p>
<p>(d) \(4x^2 - 4x + 1 = 0\)</p>
Step-by-Step Solution
Key Concept: Use the relationship R = 4r where R is circumradius and r is inradius, combined with formulas for R and r in terms of angles and sides of an isosceles triangle to derive a constraint on cos A.
<p><strong>Step 1: Set up the triangle.</strong> In isosceles △ABC with ∠A = ∠B, let ∠A = ∠B = α. Then ∠C = 180° - 2α.</p><p><strong>Step 2: Express circumradius R.</strong> Using the sine rule: R = c/(2sin C) = c/(2sin(180° - 2α)) = c/(2sin 2α)</p><p><strong>Step 3: Express inradius r.</strong> For any triangle: r = (s - a)tan(A/2) where s is semi-perimeter. Alternatively, use r = Δ/s where Δ is the area.</p><p><strong>Step 4: Use the constraint R = 4r.</strong> We have R/r = 4. Using the formula R/r = 1/(4sin(A/2)sin(B/2)sin(C/2)), we get: 1/(4sin α/2 · sin α/2 · sin(90° - α)) = 4</p><p><strong>Step 5: Simplify.</strong> This gives: 1/(4sin²(α/2) · cos α) = 4, so sin²(α/2) · cos α = 1/16</p><p><strong>Step 6: Use the half-angle formula.</strong> Since sin²(α/2) = (1 - cos α)/2, we have: ((1 - cos α)/2) · cos α = 1/16</p><p><strong>Step 7: Expand and rearrange.</strong> (1 - cos α) · cos α = 1/8, which gives: cos α - cos²α = 1/8</p><p><strong>Step 8: Convert to standard form.</strong> Rearranging: 8cos²α - 8cos α + 1 = 0. Let x = cos α, then: 8x² - 8x + 1 = 0</p><p><strong>∴ Answer:</strong> B</p>
Correct Answer: B