Statement Problem: (3)
List the angles of the triangle in order from smallest to largest.
Solution:
First, let's use the Cosine Rule to find the angle B;
[tex]\begin{gathered} b^2=a^2+c^2-2ac\cos B \\ 2ac\cos B=a^2+c^2-b^2 \\ \cos B=\frac{a^2+c^2-b^2}{2ac} \end{gathered}[/tex]
Where;
[tex]\begin{gathered} a=3,b=2.9,c=3.1 \\ \cos B=\frac{3^2+3.1^2-2.9^2}{2(3)(3.1)} \\ \cos B=\frac{9+9.61-8.41}{18.6} \\ \cos B=\frac{10.2}{18.6} \\ \cos B=0.5484 \\ B=\cos ^{-1}(0.5484) \\ B=56.74 \\ B=57^o \end{gathered}[/tex]
Also, let's use the Cosine Rule to find the angle A;
[tex]\begin{gathered} a^2=b^2+c^2-2bc\cos A \\ \cos A=\frac{b^2+c^2-a^2}{2bc} \end{gathered}[/tex][tex]\begin{gathered} \cos A=\frac{2.9^2+3.1^2-3^2}{2(2.9)(3.1)} \\ \cos A=\frac{8.41+9.61-9}{17.98} \\ \cos A=\frac{9.02}{17.68} \\ A=\cos ^{-1}(0.5102) \\ A=59.32 \\ A=59^o \end{gathered}[/tex]
Lastly, let's use the sum of angles in a triangle theorem to get the third angle, Angle C;
[tex]\begin{gathered} \angle A+\angle B+\angle C=180^o \\ \angle C=180^o-59^o-57^o \\ \angle C=64^o \end{gathered}[/tex]
Hence, the angles of the triangle from the smallest to the largest are;
[tex]\angle B,\angle A,\angle C[/tex]
Angle B,
Angle A,
Angle C
