Trigonometry & Inverse Trigonometry
Properties of Triangles
Grade 11
Question:
<p>In \(\triangle ABC\), \(\angle C = 2\angle A\) and \(AC = 2BC\), then which of the following is/are True</p>
<p>Angles A, B, C are in arithmetic progression.</p>
<p>Angles A, C, B are in arithmetic progression.</p>
<p>\(\triangle ABC\) is a right angled isosceles triangle.</p>
<p>\(BC^2 + CA^2 + AB^2 = 8R^2\) where R is the circum-radius of \(\triangle ABC\).</p>
Step-by-Step Solution
Key Concept: Use the sine rule combined with the angle relationship ∠C = 2∠A to establish ratios between sides. The constraint AC = 2BC creates a solvable system that determines the triangle's angles uniquely.
<p><strong>Step 1: Apply Sine Rule</strong></p><p>In △ABC: $\frac{BC}{\sin A} = \frac{AC}{\sin B}$</p><p>Given AC = 2BC, so: $\frac{BC}{\sin A} = \frac{2BC}{\sin B}$</p><p>Therefore: $\sin B = 2\sin A$ ... (i)</p><p><strong>Step 2: Use Angle Sum</strong></p><p>Since A + B + C = 180° and C = 2A:</p><p>$A + B + 2A = 180°$</p><p>$B = 180° - 3A$ ... (ii)</p><p><strong>Step 3: Substitute and Solve</strong></p><p>From (i): $\sin(180° - 3A) = 2\sin A$</p><p>$\sin(3A) = 2\sin A$</p><p>$3\sin A - 4\sin^3 A = 2\sin A$</p><p>$\sin A - 4\sin^3 A = 0$</p><p>$\sin A(1 - 4\sin^2 A) = 0$</p><p>Since $\sin A \neq 0$: $\sin^2 A = \frac{1}{4}$, so $\sin A = \frac{1}{2}$</p><p><strong>Step 4: Find Angles</strong></p><p>$A = 30°$, $C = 60°$, $B = 90°$</p><p>This is a right-angled triangle at B with sides in ratio 1:√3:2</p><p>∴ The triangle is a 30-60-90 right triangle, confirming specific angle measures.</p>
Correct Answer: AD