If in a triangle ABC,side a = (sqrt{3} + 1) t | vedclass

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(B) Given: $a = \sqrt{3} + 1$,$\angle B = 30^\circ$,$\angle C = 45^\circ$. Sum of angles in a triangle is $180^\circ$,so $\angle A = 180^\circ - (30^\

Vedclass · Accepted Answer

$\frac{\sqrt{3} + 1}{2} 	ext{ cm}^2$

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(B) Given: $a = \sqrt{3} + 1$,$\angle B = 30^\circ$,$\angle C = 45^\circ$. Sum of angles in a triangle is $180^\circ$,so $\angle A = 180^\circ - (30^\circ + 45^\circ) = 105^\circ$. The area of a triangle is given by $\Delta = \frac{a^2 \sin B \sin C}{2 \sin A}$. Since $\sin A = \sin 105^\circ = \sin(60^\circ + 45^\circ) = \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ = \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} + \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$. Substituting the values: $\Delta = \frac{(\sqrt{3} + 1)^2 \cdot \sin 30^\circ \cdot \sin 45^\circ}{2 \cdot \frac{\sqrt{3} + 1}{2\sqrt{2}}} = \frac{(\sqrt{3} + 1)^2 \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{2}}}{2 \cdot \frac{\sqrt{3} + 1}{2\sqrt{2}}} = \frac{(\sqrt{3} + 1)^2}{4} \cdot \frac{2\sqrt{2}}{\sqrt{3} + 1} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{3} + 1}{2} 	ext{ cm}^2$.

If in a triangle $ABC$,side $a = (\sqrt{3} + 1) \text{ cm}$,$\angle B = 30^\circ$,and $\angle C = 45^\circ$,then the area of the triangle is:

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