Trigonometry & Inverse Trigonometry
Properties of Triangles
Grade 11
Question:
<p>In triangle \(ABD\), using the sine rule, if \(BD = \sqrt{p^2+q^2}\) and \(\angle ABD = \theta\), \(\angle ADB = \alpha\), then \(AB\) equals</p>
<p>(1) \(\dfrac{(p^2+q^2)\cos\theta}{p\cos\theta + q\sin\theta}\)</p>
<p>(2) \(\dfrac{(p^2+q^2)\sin\theta}{p\sin\theta + q\cos\theta}\)</p>
<p>(3) \(\dfrac{(p^2+q^2)\sin\theta}{p\cos\theta + q\sin\theta}\)</p>
<p>(4) \(\dfrac{(p^2+q^2)\cos\theta}{p\sin\theta + q\cos\theta}\)</p>
Step-by-Step Solution
Key Concept: Apply the sine rule directly: in any triangle, the ratio of a side to the sine of its opposite angle is constant. Here, BD is opposite to ∠BAD, and AB is opposite to ∠ADB = α.
<p><strong>Step 1:</strong> Identify the angles and sides. In triangle ABD: BD = √(p² + q²) is opposite to ∠BAD, and AB is opposite to ∠ADB = α.</p><p><strong>Step 2:</strong> Find the third angle: ∠BAD = π - θ - α</p><p><strong>Step 3:</strong> Apply the sine rule: AB/sin(α) = BD/sin(∠BAD)</p><p><strong>Step 4:</strong> Substitute: AB/sin(α) = √(p² + q²)/sin(π - θ - α)</p><p><strong>Step 5:</strong> Since sin(π - θ - α) = sin(θ + α), we get:</p><p>AB = √(p² + q²) · sin(α)/sin(θ + α)</p><p>∴ Answer: C</p>
Correct Answer: C