In a triangle ABC,with usual notations,if a=4 | vedclass

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(B) Using the Law of Cosines: $c^2 = a^2 + b^2 - 2ab \cos C$ $c^2 = 4^2 + 8^2 - 2(4)(8) \cos 60^{\circ} = 16 + 64 - 64(0.5) = 80 - 32 = 48$ $c = \sqrt

Vedclass · Accepted Answer

$90^{\circ}, \sqrt{3} : 1$

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(B) Using the Law of Cosines: $c^2 = a^2 + b^2 - 2ab \cos C$ $c^2 = 4^2 + 8^2 - 2(4)(8) \cos 60^{\circ} = 16 + 64 - 64(0.5) = 80 - 32 = 48$ $c = \sqrt{48} = 4\sqrt{3}$ Now,using the Law of Sines: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ $\sin B = \frac{b \sin C}{c} = \frac{8 \sin 60^{\circ}}{4\sqrt{3}} = \frac{8(\sqrt{3}/2)}{4\sqrt{3}} = 1$ Thus,$\angle B = 90^{\circ}$ Since $\angle B = 90^{\circ}$ and $\angle C = 60^{\circ}$,then $\angle A = 180^{\circ} - 90^{\circ} - 60^{\circ} = 30^{\circ}$ The ratio $\cos A : \cos C = \cos 30^{\circ} : \cos 60^{\circ} = \frac{\sqrt{3}}{2} : \frac{1}{2} = \sqrt{3} : 1$

In a triangle $ABC$,with usual notations,if $a=4, b=8, \angle C=60^{\circ}$,then the value of $\angle B$ and the ratio $\cos A : \cos C$ respectively are,

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