JEE Mains · Maths · STD 11 - Trigonometrical equations
In a \(\Delta ABC,\frac{a}{b} = 2 + \sqrt 3 \) and \(\angle C\, = \,{60^o}.\) Then the ordered pair \((\angle A,\angle B)\) is equal to
- A \(({45^o},{75^o})\)
- B \(({105^o},{15^o})\)
- C \(({15^o},{105^o})\)
- D \(({75^o},{45^o})\)
Answer & Solution
Correct Answer
(B) \(({105^o},{15^o})\)
Step-by-step Solution
Detailed explanation
\(\frac{{\sin A}}{{\sin B}}\, = \,2\, + \,\sqrt 3 \) \(\frac{{\sin \left( {{{105}^\circ }} \right)}}{{\sin \left( {{{15}^\circ }} \right)}}\, = \,2\, + \,\sqrt 3 \) \(\frac{{\cos {{15}^\circ }}}{{\sin {{15}^\circ }}}\, = \,2\, + \,\sqrt 3 \)
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