In \(\triangle ABC\), if \(A = 45^\circ,\ B = 60^\circ\) then find the ratio of its sides.
Q.16 In \(\triangle ABC\), if \(A = 45^\circ,\ B = 60^\circ\) then find the ratio of its sides.
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Answer: \(a : b : c = 2 : \sqrt{6} : (\sqrt{3} + 1)\)
\(C = 180^\circ - 45^\circ - 60^\circ = 75^\circ\).
\(\sin 75^\circ = \sin(45^\circ + 30^\circ) = \dfrac{1}{\sqrt{2}}\cdot\dfrac{\sqrt{3}}{2} + \dfrac{1}{\sqrt{2}}\cdot\dfrac{1}{2} = \dfrac{\sqrt{3} + 1}{2\sqrt{2}}\).
By the sine rule, \(a : b : c = \sin A : \sin B : \sin C = \dfrac{1}{\sqrt{2}} : \dfrac{\sqrt{3}}{2} : \dfrac{\sqrt{3} + 1}{2\sqrt{2}}\).
Multiplying each term by \(2\sqrt{2}\): \(a : b : c = 2 : \sqrt{6} : (\sqrt{3} + 1)\).
Doubtly AI
· 28 Oct 2026
AI-verified · Oct 2026
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