In triangle ABC, $A=31.4,B=53.7,\angle C={{61.3}^{\circ }}$,how do you find the area?

Hint: The question asks for the value of area of a triangle when the 2 sides and the third angle is given . Although the area of the triangle is defined as $A=\dfrac{1}{2}\times b\times h$ , where “b” is the base of the triangle , “h” is the height of the triangle. But if we don’t know the height of the given triangle, the formula used is $A=\dfrac{1}{2}b(a\sin C)$ , where a , b are the sides of the triangle and $\angle C$ is the third angle of the triangle.

Complete step by step solution:
Now consider a triangle ABC which is not a right-angled triangle , the 2 sides are a, b and third angle is C. If we make an perpendicular to the base BC naming the line to be AD then h becomes $h=a\sin C$ the on substituting the value of h in the formula then the area A changes to $A=\dfrac{1}{2}b(a\sin C)$. Given two sides of a triangle a and b and the angle C between sides a and b (also called the inclusive angle ), the sine area formula is this formed .

 $\begin{align}
  & \sin C=\dfrac{AD}{AC}=\dfrac{f}{b} \\
 & f=b\sin C \\
 & \text{Substituting in A=}\dfrac{1}{2}base\times height(h) \\
 & \therefore \text{Area of triangle becomes } \\
 & A=\dfrac{1}{2}ab\sin C \\
\end{align}$
The values of the sides of the triangle are known to us and they are :
$A=31.4,B=53.7,\angle C={{61.3}^{\circ}}$
The sin of angle C ($\angle C$ ) is:
$\sin {{61.3}^{\circ}}=0.887$
The area of a triangle is
$A=\dfrac{1}{2}\times A\times B\times \sin \angle C$
On putting the values of the respective variable , we get
$\begin{align}
  & \Rightarrow \dfrac{1}{2}\times 31.4\times 53.7\times 0.877 \\
 & =739.38{{m}^{2}} \\
\end{align}$
$\therefore $ The area of the triangle is $739.38{{m}^{2}}$.

Note: The value of sine of an angle will always be less than 1 . The formula given above can be applied to any type of triangle being acute , obtuse , right angle for finding the area of the triangle. This works when 2 sides and the third angle of a triangle are given.