Question

In \(\triangle ABC\), if \(A = 30^\circ\) and

\[ \frac{b}{(\sqrt{3}+1)^2 + 2(\sqrt{2} - 1)}, \quad \frac{c}{(\sqrt{3}+1)^2 - 2(\sqrt{2} - 1)}, \]

Then find the angle \(B\).

• A. \(60^\circ\)
• B. \(97.5^\circ\)
• C. \(75^\circ\)
• D. \(52.5^\circ\)

Hint:

When dealing with radical expressions in triangles, use Law of Sines or Law of Cosines carefully and approximate if necessary to find angles.

Answer 1

The Correct Option is B

Given the expressions involve side ratios in terms of radical expressions, we consider the Law of Cosines or Law of Sines to find \(B\). Calculate the denominators: \[ (\sqrt{3} + 1)^2 = (\sqrt{3})^2 + 2 . \sqrt{3} . 1 + 1^2 = 3 + 2\sqrt{3} + 1 = 4 + 2\sqrt{3}, \] and \[ 2(\sqrt{2} - 1) = 2\sqrt{2} - 2. \] So, \[ \frac{b}{4 + 2\sqrt{3} + 2\sqrt{2} - 2} = \frac{b}{(4 - 2) + 2\sqrt{3} + 2\sqrt{2}} = \frac{b}{2 + 2\sqrt{3} + 2\sqrt{2}}, \] and \[ \frac{c}{4 + 2\sqrt{3} - 2\sqrt{2} + 2} = \frac{c}{6 + 2\sqrt{3} - 2\sqrt{2}}. \] Using the ratio of sides and the Law of Cosines, one obtains angle \(B \approx 97.5^\circ\) after simplification (detailed trigonometric steps depend on algebraic manipulation and approximations).