Trigonometry & Inverse Trigonometry
Sine Rule in Triangles
Grade 11
Question:
<p>With the usual notation, in triangle ABC, if $\angle A + \angle B = 120°$, $a = \sqrt{3} + 1$ and $b = \sqrt{3} - 1$, then the ratio $\angle A : \angle B$ is</p>
<p>(a) 7 : 1</p>
<p>(b) 3 : 1</p>
<p>(c) 9 : 7</p>
<p>(d) 5 : 3</p>
Step-by-Step Solution
Key Concept: Use the Sine Rule to relate the sides to angles, then apply the constraint that angles sum to 120° to solve for individual angles.
<p>Given: $a = \sqrt{3} + 1$, $b = \sqrt{3} - 1$, and $\angle A + \angle B = 120°$</p><p>By the Sine Rule: $\frac{a}{\sin A} = \frac{b}{\sin B}$</p><p>$\frac{\sqrt{3}+1}{\sin A} = \frac{\sqrt{3}-1}{\sin B}$</p><p>$\frac{\sin A}{\sin B} = \frac{\sqrt{3}+1}{\sqrt{3}-1}$</p><p>Rationalizing: $\frac{\sin A}{\sin B} = \frac{(\sqrt{3}+1)^2}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{3 + 2\sqrt{3} + 1}{3-1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$</p><p>Since $\angle A + \angle B = 120°$, we have $\angle B = 120° - \angle A$</p><p>$\frac{\sin A}{\sin(120° - A)} = 2 + \sqrt{3}$</p><p>Using $\sin(120° - A) = \sin 120° \cos A - \cos 120° \sin A = \frac{\sqrt{3}}{2}\cos A + \frac{1}{2}\sin A$</p><p>Solving this equation yields $\angle A = 105°$ and $\angle B = 15°$</p><p>Therefore, $\angle A : \angle B = 105° : 15° = 7 : 1$</p><p>∴ Answer is (a)</p>
Correct Answer: A