Question

Find the area of the triangle with the given measurements. Round the solution to the nearest hundredth if necessary.

B = 104°, a = 11 cm, c = 18 cm

Answer 1

`\bf \textit{Law of Cosines}\\ \quad \\ c^2 = {{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)\implies c = \sqrt{{{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)}\\\\ -----------------------------\\\\ b = \sqrt{{{ a}}^2+{{ c}}^2-(2{{ a}}{{ c}})cos(B)} \\\\\\ b=\sqrt{{{ 11}}^2+{{ 18}}^2-2(11\cdot 18)cos(104^o)} \\\\\\ b\approx 23.25512998582180400481\implies b\approx 23.26`

so, "b" rounded up is 23.26, now, we know a = 11 and c = 18

well, to get the area then, let us use Heron's formula


`\bf \textit{Heron's Area formula}\\\\ A=√(s(s-a)(s-b)(s-c))\qquad \begin{cases} a=11\\ c=18\\ b\approx 23.26\\\\ s=\cfrac{a+b+c}{2} \end{cases}`