Area of Triangles

Find the area of triangle ABC with $\angle A=40\degree,\ b=4cm,\ and\ c=8cm$   

 $A=\frac{1}{2}bc\sin A$  

 $A=\frac{1}{2}\left(8\right)\left(4\right)\sin40$  

 $A=10.3cm^2$  

The area of triangle XYZ is $36cm^2$ . If side x=12 and y=8 find the measurements of the included angles Z.

 $A=\frac{1}{2}xy\sin Z$  

 $36=\frac{1}{2}\left(12\right)\left(8\right)\sin Z$  

 $36=48\sin Z$  

 $\frac{3}{4}=\sin Z$  (since positive sin the angles  can be found in the first and second quadrant

Z= $48.6\degree\ or\ 180-48.6=131.4\degree$  

Find the area of triangle XYZ if x=2, y=7, and z=8

 $s=\frac{x+y+z}{2}$  First solve for s.  

 $s=\frac{2+7+8}{2}$  so s= 8.5

Substitute values into formula and solve 

 $A=\sqrt{s\left(s-x\right)\left(s-y\right)\left(s-z\right)}$  

 $A=\sqrt{8.5\left(8.5-2\right)\left(8.5-7\right)\left(8.5-8\right)}$  

 $A=\sqrt{41.4375},\ A=6.44$  units squared