1) We know that the sum of the interior angles within a triangle is 180º
So we can state that :
∠X = 80º , ∠V=32º and ∠W = 180º-(80+32), ∠W=68º
2) The area of a triangle can be found by the Heron formula, but before that, we need to find out the other legs. Let's sketch this and use the Law of Sines:
The Law of Sines:
[tex]\begin{gathered} \frac{v}{\sin(V)}=\frac{w}{\sin(W)}=\frac{x}{\sin(X)} \\ \frac{v}{\sin(32)}=\frac{680}{\sin(68)} \\ 680\text{ sin(32) = vsin(68)} \\ \frac{680\text{ }\sin(32_{}}{\sin(68)}=v \\ v\text{ =}388.65 \\ \\ \end{gathered}[/tex]Now let's proceed to find the length of side x:
[tex]\begin{gathered} \frac{w}{\sin(W)}=\frac{x}{\sin(X)} \\ \frac{680}{\sin(68)}=\frac{x}{\sin (80)} \\ x\sin (68)=680\cdot\sin (80) \\ x=\frac{680\cdot\sin (80)}{\sin (68)} \\ x\approx722.26 \end{gathered}[/tex]Now we can add the three sides and find out the Perimeter:
2p: 722.26 +388.65 +680
2p=1790.91
And the semi perimeter is p =2p/2, p =895.455.
2.2) Finally we can find out the area using the Heron Formula:
[tex]\begin{gathered} A=\sqrt[]{p(p-v)(p-w)(p-x)} \\ A=\sqrt[]{895.455(895.455-388.65)(895.455-722.455)(895.455-680)} \\ A=130059.9757\approx130060 \end{gathered}[/tex]3) Hence, the area is 130,060 cm²
