Q58 / 150 · MHT CET 2026 (16 April Shift 1) · Mathematics
In a triangle ABC, with usual notations $a = \sqrt{3} + 1$, $b = \sqrt{3} - 1$ and $\angle C = 60^\circ$ then the values of $\angle A$ and $\angle B$ respectively are
- Correct answerA$105^\circ, 15^\circ$
- B$100^\circ, 20^\circ$
- C$90^\circ, 30^\circ$
- D$110^\circ, 10^\circ$
Solution
Using Napier's Analogy:
$\tan\left(\dfrac{A - B}{2}\right) = \dfrac{a - b}{a + b} \cot\left(\dfrac{C}{2}\right)$
Substituting the given values $a = \sqrt{3} + 1$, $b = \sqrt{3} - 1$, and $C = 60^\circ$:
$a - b = 2$
$a + b = 2\sqrt{3}$
$\cot\left(\dfrac{C}{2}\right) = \cot 30^\circ = \sqrt{3}$
$\tan\left(\dfrac{A - B}{2}\right) = \dfrac{2}{2\sqrt{3}} \times \sqrt{3} = 1$
$\dfrac{A - B}{2} = 45^\circ \Rightarrow A - B = 90^\circ$
In $\triangle ABC$, the sum of angles is $180^\circ$:
$A + B + C = 180^\circ \Rightarrow A + B = 120^\circ$
Solving $A - B = 90^\circ$ and $A + B = 120^\circ$:
$2A = 210^\circ \Rightarrow A = 105^\circ$
$2B = 30^\circ \Rightarrow B = 15^\circ$
Answer: $105^\circ, 15^\circ$
$\tan\left(\dfrac{A - B}{2}\right) = \dfrac{a - b}{a + b} \cot\left(\dfrac{C}{2}\right)$
Substituting the given values $a = \sqrt{3} + 1$, $b = \sqrt{3} - 1$, and $C = 60^\circ$:
$a - b = 2$
$a + b = 2\sqrt{3}$
$\cot\left(\dfrac{C}{2}\right) = \cot 30^\circ = \sqrt{3}$
$\tan\left(\dfrac{A - B}{2}\right) = \dfrac{2}{2\sqrt{3}} \times \sqrt{3} = 1$
$\dfrac{A - B}{2} = 45^\circ \Rightarrow A - B = 90^\circ$
In $\triangle ABC$, the sum of angles is $180^\circ$:
$A + B + C = 180^\circ \Rightarrow A + B = 120^\circ$
Solving $A - B = 90^\circ$ and $A + B = 120^\circ$:
$2A = 210^\circ \Rightarrow A = 105^\circ$
$2B = 30^\circ \Rightarrow B = 15^\circ$
Answer: $105^\circ, 15^\circ$
Hence, the correct answer is option: 'A'.