If in a Delta ABC,angle A = 45^circ,angle C = | vedclass

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(B) Given: $\angle A = 45^\circ, \angle C = 60^\circ$.Since the sum of angles in a triangle is $180^\circ$,$\angle B = 180^\circ - (45^\circ + 60^\cir

Vedclass · Accepted Answer

$2b$

Vedclass · Answer

(B) Given: $\angle A = 45^\circ, \angle C = 60^\circ$.Since the sum of angles in a triangle is $180^\circ$,$\angle B = 180^\circ - (45^\circ + 60^\circ) = 75^\circ$.Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$,so $a = k\sin A, c = k\sin C, b = k\sin B$.We need to evaluate $a + c\sqrt{2} = k\sin 45^\circ + k\sqrt{2}\sin 60^\circ$.$a + c\sqrt{2} = k\left(\frac{1}{\sqrt{2}}\right) + k\sqrt{2}\left(\frac{\sqrt{3}}{2}\right) = k\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{6}}{2}\right) = k\left(\frac{\sqrt{2} + \sqrt{6}}{2}\right)$.Note that $\sin 75^\circ = \sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}} = \frac{\sqrt{6} + \sqrt{2}}{4}$.Thus,$a + c\sqrt{2} = k(2 \sin 75^\circ) = 2k \sin B = 2b$.

If in a $\Delta ABC$,$\angle A = 45^\circ$,$\angle C = 60^\circ$,then $a + c\sqrt{2} = $

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