Trigonometry & Inverse Trigonometry
Properties of triangles
Grade 11
Question:
<p>In a triangle \( PQR \), \( \angle R = \dfrac{\pi}{2} \). If \( \tan\left(\dfrac{P}{2}\right) \) and \( \tan\left(\dfrac{Q}{2}\right) \) are roots of \( ax^2 + bx + c = 0 \) (where \( a \neq 0 \)), then which of the following is true?</p>
<p>\( b = a + c \)</p>
<p>\( c = a + b \)</p>
<p>\( a = b + c \)</p>
<p>\( b = c \)</p>
Step-by-Step Solution
Key Concept: Since ∠R = π/2 in triangle PQR, we have P + Q = π/2. Use the constraint P + Q = π/2 to relate tan(P/2) and tan(Q/2), then apply Vieta's formulas to the quadratic equation.
<p><strong>Step 1:</strong> Use the triangle angle constraint. Since ∠R = π/2, we have P + Q = π/2, which gives P/2 + Q/2 = π/4.</p><p><strong>Step 2:</strong> Apply the tangent addition formula. Since P/2 + Q/2 = π/4:</p><p>tan(P/2 + Q/2) = tan(π/4) = 1</p><p>Therefore: $\frac{\tan(P/2) + \tan(Q/2)}{1 - \tan(P/2)\tan(Q/2)} = 1$</p><p><strong>Step 3:</strong> Simplify the equation:</p><p>$\tan(P/2) + \tan(Q/2) = 1 - \tan(P/2)\tan(Q/2)$</p><p>$\tan(P/2) + \tan(Q/2) + \tan(P/2)\tan(Q/2) = 1$</p><p><strong>Step 4:</strong> Apply Vieta's formulas. Let α = tan(P/2) and β = tan(Q/2) be roots of ax² + bx + c = 0:</p><p>Sum: α + β = -b/a</p><p>Product: αβ = c/a</p><p><strong>Step 5:</strong> Substitute into our constraint:</p><p>$-\frac{b}{a} + \frac{c}{a} = 1$</p><p>$-b + c = a$</p><p>∴ <strong>b + a = c</strong> (or equivalently: a + b = c)</p>
Correct Answer: B