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Question

In a triangle ABC , if $$\angle A = 45^0 , \angle  C =60^0,$$ the a+c is equal to 


A
(2+22)k.
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B
(2+32)k.
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C
(2+33)k.
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D
(3+32)k.
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Solution

The correct option is B $$\left(\dfrac{\sqrt{2}+\sqrt{3}}{2}\right)k.$$
$$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin c}=k$$        $$\left[ A=45°,B=75°,C=60° \right] $$
$$\Rightarrow a=k\sin A=k\sin 45°=\dfrac{k}{\sqrt{2}}$$
     $$c=k\sin C=k\sin 60°=\dfrac{\sqrt{3}}{2}$$
     $$b=k\sin B=k\sin 75°$$
$$\Rightarrow a+c=k\left(\dfrac{1}{\sqrt{3}}+\dfrac{\sqrt{3}}{2}\right)=\left(\dfrac{\sqrt{2}+\sqrt{3}}{2}\right)k$$
Hence, the answer is $$\left(\dfrac{\sqrt{2}+\sqrt{3}}{2}\right)k.$$

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