MHT CET · Maths · Properties of Triangles
In a triangle ABC , with usual notations if \(a=4, \mathrm{~b}=8, \angle \mathrm{C}=60^{\circ}\), then the value of \(\angle \mathrm{B}\) and the ratio \(\cos \mathrm{A}: \cos \mathrm{C}\) respectively are,
- A \(\frac{\pi}{4},~ 1: \sqrt{3}\)
- B \(\frac{\pi}{2},~ \sqrt{3}: 1\)
- C \(\frac{\pi}{2},~ 2: \sqrt{3}\)
- D \(\frac{\pi}{6},~ \sqrt{3}: 2\)
Answer & Solution
Correct Answer
(B) \(\frac{\pi}{2},~ \sqrt{3}: 1\)
Step-by-step Solution
Detailed explanation
\(c^2 = a^2 + b^2 - 2ab \cos C = 4^2 + 8^2 - 2(4)(8) \cos 60^\circ = 16 + 64 - 64(0.5) = 80 - 32 = 48\) \(c = \sqrt{48} = 4\sqrt{3}\)
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