Trigonometry & Inverse Trigonometry
Sine Rule in Triangles
Grade 11
Question:
<p>Let <i>ABC</i> be a triangle such that ∠<i>A</i> = 45°, ∠<i>B</i> = 75°, then \(a + c\sqrt{2}\) is equal to</p>
<p>(a) 0</p>
<p>(b) <i>b</i></p>
<p>(c) 2<i>b</i></p>
<p>(d) -<i>b</i></p>
Step-by-Step Solution
Key Concept: Use the sine rule and compute trigonometric values for special angles.
<p>Since ∠<i>A</i> = 45°, ∠<i>B</i> = 75°, we have ∠<i>C</i> = 180° - 45° - 75° = 60°.</p><p>Using sine rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$</p><p>$\frac{a}{\sin 45°} = \frac{b}{\sin 75°} = \frac{c}{\sin 60°}$</p><p>From this, $a = \frac{b \sin 45°}{\sin 75°}$ and $c = \frac{b \sin 60°}{\sin 75°}$</p><p>Computing: $a + c\sqrt{2} = \frac{b(\sin 45° + \sqrt{2}\sin 60°)}{\sin 75°} = 2b$</p>
Correct Answer: C