In triangle ABC,if A=75^{circ} and B=45^{circ | vedclass

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Question

(C) Given in $	riangle ABC$,$A=75^{\circ}$ and $B=45^{\circ}$.Using the angle sum property of a triangle,$C = 180^{\circ} - (75^{\circ} + 45^{\circ})

Vedclass · Accepted Answer

$2a$

Vedclass · Answer

(C) Given in $	riangle ABC$,$A=75^{\circ}$ and $B=45^{\circ}$.Using the angle sum property of a triangle,$C = 180^{\circ} - (75^{\circ} + 45^{\circ}) = 60^{\circ}$.Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$.Thus,$b = k \sin 45^{\circ} = k \frac{1}{\sqrt{2}}$ and $c = k \sin 60^{\circ} = k \frac{\sqrt{3}}{2}$.Also,$a = k \sin 75^{\circ} = k \sin(45^{\circ}+30^{\circ}) = k \left( \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} ight) = k \frac{\sqrt{3}+1}{2\sqrt{2}}$.We need to find $b + c\sqrt{2}$.$b + c\sqrt{2} = k \frac{1}{\sqrt{2}} + k \frac{\sqrt{3}}{2} \cdot \sqrt{2} = k \left( \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{\sqrt{2}} ight) = k \frac{\sqrt{3}+1}{\sqrt{2}}$.Since $a = k \frac{\sqrt{3}+1}{2\sqrt{2}}$,we have $k = \frac{2\sqrt{2}a}{\sqrt{3}+1}$.Substituting $k$ back,$b + c\sqrt{2} = \left( \frac{2\sqrt{2}a}{\sqrt{3}+1} ight) \left( \frac{\sqrt{3}+1}{\sqrt{2}} ight) = 2a$.

In $\triangle ABC$,if $A=75^{\circ}$ and $B=45^{\circ}$,then the value of $b+c\sqrt{2}$ is:

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