Triangle Area #2

We can use the Heron's formula to solve this:

So in this case, we have

$A=\sqrt{25(25-12)(25-16)(25-22)}$

Which simplifies to

$A=\sqrt{25(13)(9)(3)}$

Which equals to

$A=\sqrt{8875}$

$=93.675$ To (3dp)

By cosine rule, we have

$22^2=12^2+16^2-2(12)(16)(\cos A)$

$A \approx 102.636$

So the area is $\dfrac{1}{2}(12)(16)(\sin 102.636) \approx \boxed{93.675~m^2}$