Trigonometry & Inverse Trigonometry
Properties of triangles
Grade 11
Question:
<p>In \(\triangle ABC\), \(a=11\) and \(\sin A=\dfrac{3}{7}\) where '<em>a</em>' is the side opposite to \(\angle A\) and \(0 < A < \dfrac{\pi}{2}\). If the side length of \(\triangle ABC\) are 11, <em>b</em>, <em>c</em> where '<em>b</em>' is the largest possible side of \(\triangle ABC\), then:</p>
<p>(a) circumradius \(R\) of \(\triangle ABC\) is equal to \(\dfrac{77}{6}\)</p>
<p>(b) inradius \(r\) of \(\triangle ABC\) is equal to \((11)(\sqrt{2})\left(\dfrac{\sqrt{5}-\sqrt{2}}{3}\right)\)</p>
<p>(c) area of \(\triangle ABC\) is equal to \(\dfrac{121\sqrt{10}}{3}\)</p>
<p>(d) the value of \(\sin 2A+\sin 2B+\sin 2C\) is equal to \(\dfrac{24\sqrt{10}}{7}\)</p>
Step-by-Step Solution
Key Concept: Since a=11 and sin A=3/7 with 0<A<π/2, use the sine rule a/sin A = 2R to find circumradius R. The largest side b occurs when angle B=π/2 (right angle at B), making b the hypotenuse opposite the largest angle.
<p><strong>Step 1: Find the circumradius using sine rule</strong></p><p>From a/sin A = 2R, we have: 11/(3/7) = 2R</p><p>Therefore: 2R = 77/3, so R = 77/6</p><p><strong>Step 2: Determine that b is the largest side</strong></p><p>Since b is stated as the largest side and a=11 is given, angle B must be the largest angle. For b to be maximum while maintaining triangle validity, angle B = π/2 (right angle).</p><p><strong>Step 3: Find b using sine rule when B = π/2</strong></p><p>When B = π/2: sin B = 1, so b = 2R·sin B = (77/6)·1 = 77/6</p><p><strong>Step 4: Find angle C and side c</strong></p><p>Since A + B + C = π and B = π/2:</p><p>C = π/2 - A, so sin C = sin(π/2 - A) = cos A</p><p>cos A = √(1 - sin²A) = √(1 - 9/49) = √(40/49) = 2√10/7</p><p>c = 2R·sin C = (77/6)·(2√10/7) = (77·2√10)/(6·7) = 11√10/3</p><p><strong>Step 5: Verify the conditions</strong></p><p>Check: b = 77/6 ≈ 12.83 is indeed larger than a = 11 ✓</p><p>Check: c = 11√10/3 ≈ 11.6 means b > c > a ✓</p><p>∴ Answer: ACD</p>
Correct Answer: ACD