Question

In a triangle \(ABC\), \[ \frac{a}{b}=2+\sqrt{3} \] 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:

In a triangle, use the sine rule: \[ \frac{a}{b}=\frac{\sin A}{\sin B} \] and the angle sum property: \[ A+B+C=180^\circ \] to relate sides and angles.

Answer 1

The Correct Option is C

Step 1: Use sine rule in triangle \(ABC\).
By sine rule, \[ \frac{a}{b}=\frac{\sin A}{\sin B} \] Given, \[ \frac{a}{b}=2+\sqrt{3} \] Therefore, \[ \frac{\sin A}{\sin B}=2+\sqrt{3} \]

Step 2: Use angle sum property.
In triangle \(ABC\), \[ A+B+C=180^\circ \] Given, \[ C=60^\circ \] So, \[ A+B=120^\circ \] Thus, \[ B=120^\circ-A \]

Step 3: Substitute \(B=120^\circ-A\).
\[ \frac{\sin A}{\sin(120^\circ-A)}=2+\sqrt{3} \] Now check the given options.
For option (3), \[ A=105^\circ \] Then, \[ B=120^\circ-105^\circ=15^\circ \] So, \[ \frac{\sin A}{\sin B} = \frac{\sin105^\circ}{\sin15^\circ} \]

Step 4: Evaluate the ratio.
We know, \[ \sin105^\circ=\sin(60^\circ+45^\circ) \] \[ =\sin60^\circ\cos45^\circ+\cos60^\circ\sin45^\circ \] \[ =\frac{\sqrt3}{2}\cdot\frac{\sqrt2}{2} + \frac12\cdot\frac{\sqrt2}{2} \] \[ =\frac{\sqrt6+\sqrt2}{4} \] Also, \[ \sin15^\circ=\sin(45^\circ-30^\circ) \] \[ =\sin45^\circ\cos30^\circ-\cos45^\circ\sin30^\circ \] \[ =\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2} - \frac{\sqrt2}{2}\cdot\frac12 \] \[ =\frac{\sqrt6-\sqrt2}{4} \] Therefore, \[ \frac{\sin105^\circ}{\sin15^\circ} = \frac{\sqrt6+\sqrt2}{\sqrt6-\sqrt2} \] Rationalizing, \[ = \frac{(\sqrt6+\sqrt2)^2}{6-2} \] \[ = \frac{6+2+2\sqrt{12}}{4} \] \[ = \frac{8+4\sqrt3}{4} \] \[ =2+\sqrt3 \] This matches the given value of \(\frac{a}{b}\).

Step 5: Final conclusion.
Hence, \[ \angle A=105^\circ \] Therefore, \[ \boxed{105^\circ} \] which corresponds to option (3).