Answer:
Angles: 40.9°, 99.05°, 40.05°
Step-by-step explanation:
Triangles
When we are given the lengths of the 3 sides of a triangle, we can easily compute all the internal angles by using the cosine's law or formula. Being x,y and z the sides of a triangle, and \alpha, \beta , \gamma the three opposite angles respectively, then
[tex]x^2=y^2+z^2-2yzcos\alpha[/tex]
[tex]y^2=x^2+z^2-2xzcos\beta[/tex]
[tex]z^2=x^2+y^2-2xycos\gamma[/tex]
We have x=5.9, y=8.9, z=5.8, then from
[tex]x^2=y^2+z^2-2yzcos\alpha[/tex]
We solve for [tex]\alpha[/tex]
[tex]\displaystyle cos\alpha=\frac{y^2+z^2-x^2}{2yz}[/tex]
[tex]\displaystyle cos\alpha=\frac{8.9^2+5.8^2-5.9^2}{2\times 8.9\times 5.8}[/tex]
[tex]cos \alpha=0.756[/tex]
[tex]\alpha=40.9 ^o[/tex]
Similarly
[tex]\displaystyle cos\beta=\frac{x^2+z^2-y^2}{2xz}[/tex]
[tex]\displaystyle cos\beta=\frac{5.9^2+5.8^2-8.9^2}{2\times 5.9\times 5.8}[/tex]
[tex]cos\beta=-0.157[/tex]
[tex]\beta=99.05^o[/tex]
Finally
[tex]\displaystyle cos\gamma=\frac{5.9^2+8.9^2-5.8^2}{2\times 5.9\times 8.9}[/tex]
[tex]cos\gamma=0.765[/tex]
[tex]\gamma=40.05^o[/tex]
