ACCORDING TO QUES:
In a \[\vartriangle ABC,\text{ }a\text{ }=\text{ }5,\text{ }b\text{ }=\text{ }6\text{ }and\text{ }C\text{ }=\text{ }{{60}^{o}}\]
By using the formula,
Area of\[\vartriangle ABC\text{ }=\text{ }1/2\text{ }ab\text{ }sin\text{ }\theta \]
Where, lengths of the sides of a triangle are\[a\text{ }and\text{ }b\]
And, the angle between sides is\[\theta \]
So,
\[Area\text{ }of\text{ }\vartriangle ABC\text{ }=\text{ }1/2\text{ }ab\text{ }sin\text{ }\theta \]
\[=\text{ }1/2\text{ }\times \text{ }5\text{ }\times \text{ }6\text{ }\times \text{ }sin\text{ }{{60}^{o}}\]
\[=\text{ }30/2\text{ }\times ~\surd 3/2\]
\[=\text{ }\left( 15\surd 3 \right)/2\text{ }sq.\text{ }units\]
Hence proved