Step 1:
First, we need to determine if the triangle is solvable or not by applying the triangle inequality theorem which states that the sum of any two sides of a triangle must be greater than the measure of the third side
In our case,
[tex]\begin{gathered} a+b>c \\ a+c>b \\ b+c>a \end{gathered}[/tex]
Therefore, the triangle is solvable
Step 2:
The given sides of the triangle are a = 6.3, b = 9.3, c =8.3
We need to find the three angles of the triangles labelled above. In order to do this, we need to apply the cosine rule,
To calculate the first angle α,
[tex]\begin{gathered} \cos \alpha=\frac{b^2+c^2-a^2}{2bc} \\ \cos \alpha=\frac{9.3^2+8.3^2-6.3^2}{2\times9.3\times8.3} \\ \cos \alpha=\frac{86.49^{}+68.89^{}-36.69}{117.18} \\ \cos \alpha=\frac{115.69^{}}{154.38} \\ \cos \alpha=0.7494 \\ \alpha=41.5^0 \end{gathered}[/tex]
To calculate the second angle β
[tex]\begin{gathered} \cos \beta=\frac{a^2+c^2-b^2^{}}{2ac} \\ \cos \beta=\frac{6.3^2+8.3^2-9.3^2}{2\times6.3\times8.3} \\ \cos \beta=\frac{39.69^{}+68.89^{}-86.49^{}}{2\times6.3\times8.3} \\ \cos \beta=\frac{22.09^{}}{104.58} \\ \cos \beta=0.2112 \\ \beta=\cos ^{-1}(0.2112) \\ \beta=77.8^0 \end{gathered}[/tex]
To calculate the third angle,
[tex]\begin{gathered} \gamma=\text{ 180 - (41.5+77.8})\text{ (sum of angles in a triangle is 180deg.)} \\ \gamma=60.7^0 \end{gathered}[/tex]
Therefore, the measures of the angles of the triangle are:
[tex]\alpha=41.5^0,\text{ }\beta=77.8^0,\text{ }\gamma=60.7^0[/tex]
