Answer
a) $θ\approx 130º$
b) $A\approx45$ units$^2$
Work Step by Step
a) Here one can use the law of cosines:
$20^2=13^2+9^2-2(13)(9)\cdot $cos$(θ)$
$400=169+81-234\cdot $cos$(θ)$
$400-169-81=-234\cdot $cos$(θ)$
cos$(θ)=\frac{150}{-234} $
$θ=$cos$^{-1}(\frac{150}{-234})\approx 130º$
b) Heron's formula should be used, but $s$ (the semi-perimeter) must be found first:
$s=\frac{1}{2}(9+13+20)$
$s=\frac{1}{2}(42)=21$
Now, the area can be found:
$A=\sqrt{21(21-20)(21-13)(21-9)}$
$A=\sqrt{21(1)(8)(12)}$
$A=\sqrt{2016}\approx45$ units$^2$
