Find the area of the triangle ` DeltaA B C` in which `a=1,\ b=2\ a n d\ /_c=60^0dot`

Explanation

`a=1, ~b=2` and `angle C=60^{circ}`

By Cosine law,

`cos (c)=frac{a^{2}+b^{2}-c^{2}}{2 a b}`

`frac{1}{2}-frac{1^{2}+2^{2}-c^{2}}{2 times 1 times 2}`

`2=1+4 cdot c^{2}`

`c^{2}=5 cdot 2`

`c^{2}=3`

`c=sqrt{3}`

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