In a Delta ABC,if angle C = 30^circ,a = 47 te | vedclass

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(D) Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$ Given $\angle C = 30^\circ$,$a = 47$,$b = 94$: $\cos 30^\circ = \frac{47^2 + 94^2

Vedclass · Accepted Answer

Obtuse angled

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(D) Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$ Given $\angle C = 30^\circ$,$a = 47$,$b = 94$: $\cos 30^\circ = \frac{47^2 + 94^2 - c^2}{2 	imes 47 	imes 94}$ $\frac{\sqrt{3}}{2} = \frac{2209 + 8836 - c^2}{8836}$ $4418\sqrt{3} = 11045 - c^2$ $c^2 = 11045 - 7653.58 \approx 3391.42$ $c \approx 58.24 	ext{ cm}$ Now,use the Law of Sines: $\frac{a}{\sin A} = \frac{c}{\sin C}$ $\sin A = \frac{a \sin C}{c} = \frac{47 	imes 0.5}{58.24} \approx 0.4035$ $A \approx 23.8^\circ$ $\angle B = 180^\circ - (30^\circ + 23.8^\circ) = 126.2^\circ$ Since $\angle B > 90^\circ$,the triangle is an obtuse angled triangle.

In a $\Delta ABC$,if $\angle C = 30^\circ$,$a = 47 \text{ cm}$ and $b = 94 \text{ cm}$,then the triangle is

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