In a triangle ABC,with usual notations,if m a | vedclass

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(C) Given that: $m \angle A = 45^{\circ}, m \angle B = 75^{\circ}$.Since the sum of angles in a triangle is $180^{\circ}$,$m \angle C = 180^{\circ} -

Vedclass · Accepted Answer

$2b$

Vedclass · Answer

(C) Given that: $m \angle A = 45^{\circ}, m \angle B = 75^{\circ}$.Since the sum of angles in a triangle is $180^{\circ}$,$m \angle C = 180^{\circ} - (45^{\circ} + 75^{\circ}) = 60^{\circ}$.Using the Sine Rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$.Thus,$a = k \sin 45^{\circ} = \frac{k}{\sqrt{2}}$,$b = k \sin 75^{\circ} = k \sin(45^{\circ} + 30^{\circ}) = k \left( \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \right) = \frac{k(\sqrt{3} + 1)}{2\sqrt{2}}$,and $c = k \sin 60^{\circ} = \frac{k\sqrt{3}}{2}$.We need to find $a + c\sqrt{2}$:$a + c\sqrt{2} = \frac{k}{\sqrt{2}} + \frac{k\sqrt{3}}{2} \cdot \sqrt{2} = \frac{k}{\sqrt{2}} + \frac{k\sqrt{3}}{\sqrt{2}} = \frac{k(1 + \sqrt{3})}{\sqrt{2}}$.From the expression for $b$,we have $b = \frac{k(\sqrt{3} + 1)}{2\sqrt{2}}$,which implies $2b = \frac{k(\sqrt{3} + 1)}{\sqrt{2}}$.Therefore,$a + c\sqrt{2} = 2b$.

In a triangle $ABC$,with usual notations,if $m \angle A = 45^{\circ}$ and $m \angle B = 75^{\circ}$,then $a + c \sqrt{2}$ is equal to:

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