Question

In a triangle $ABC$, $\dfrac{a}{b} = 2 + \sqrt{3}$ and $\angle C = 60^\circ$. Then the measure of $\angle A$ is

• A. $95^\circ$
• B. $65^\circ$
• C. $105^\circ$
• D. $115^\circ$

Hint:

Use sine rule and known angle approximations to back-calculate angles from given ratios.

Answer 1

The Correct Option is C

Using the Law of Sines: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$
Given $\dfrac{a}{b} = 2 + \sqrt{3}$ and $\angle C = 60^\circ$
Let’s use Law of Sines: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} \Rightarrow \dfrac{a}{b} = \dfrac{\sin A}{\sin B}$
Also, from angle sum in triangle: $A + B + C = 180^\circ \Rightarrow A + B = 120^\circ$
Try $A = 105^\circ$, then $B = 15^\circ$
Now check: $\dfrac{\sin 105^\circ}{\sin 15^\circ} \approx \dfrac{0.9659}{0.2588} \approx 3.73$
$2 + \sqrt{3} \approx 2 + 1.732 = 3.732$ — matches
Hence, $\angle A = 105^\circ$