Trigonometry & Inverse Trigonometry
Properties of Triangles and Sine Rule
Grade 11
Question:
<p>In triangle ABC, if ∠A = 30°, b = 10 and a = x, then the values of x for which there are 2 possible triangles is given by (All symbols have usual meaning in a triangle)</p>
<p>(a) 5 < x < 10</p>
<p>(b) x < 5/2</p>
<p>(c) 5/3 < x < 10</p>
<p>(d) 5/2 < x < 10</p>
Step-by-Step Solution
Key Concept: For two possible triangles in SSA case, the given side a must satisfy: (altitude from C) < a < b
<p><strong>Step 1:</strong> The altitude from C to side AB has length \(10\sin 30° = 5\).</p><p><strong>Step 2:</strong> For two possible triangles to exist, x must be greater than the altitude and less than b. So \(5 < x < 10\).</p><p><strong>Step 3:</strong> If \(x > 10\), point B would come to the left of A and ∠A would be obtuse.</p><p><strong>Step 4:</strong> Using the cosine rule: \(\cos 60° = \frac{100 + c^2 - x^2}{2(10)(c)}\)</p><p><strong>Step 5:</strong> This gives \(c^2 - 10\sqrt{3}c + (100 - x^2) = 0\), which yields \(c = \frac{10\sqrt{3} \pm \sqrt{4x^2 - 100}}{2}\)</p><p><strong>Step 6:</strong> For two distinct positive values of c, we need \(300 > 4x^2 - 100\), which gives \(x^2 < 100\), so \(5 < x < 10\).</p><p>∴ Answer is (a).</p>
Correct Answer: A