Solution
Given:
B = 100 degrees
c = 5
a = 2
We are interested in getting b, C and A
In any Given triangle ABC with sides a,b,c facing angles A,B,C respectively,
[tex]\begin{gathered} b^2=a^2+c^2-2acCosB----------(co\sin e\text{ rule)} \\ \end{gathered}[/tex]
Thus,
[tex]\begin{gathered} b^2=2^2+5^2-2(2)(5)Cos100 \\ b^2=4+25-20(-0.1736) \\ b^2=29+3.47 \\ b^2=32.47 \\ b=\sqrt[]{32.47} \\ b=5.6982 \\ b=5.698\text{ (nearest thousandth)} \end{gathered}[/tex]
To calculate angle A:
[tex]\begin{gathered} \text{ Using sine rule, } \\ \frac{a}{\sin\text{ A}}=\frac{b}{\sin B} \\ \frac{2}{\sin\text{ A}}=\frac{5.6982}{\sin100} \\ \\ 5.6982\text{ x sinA=2 x sin 100} \\ \sin A=\frac{2\text{ x sin 100}}{5.6982} \\ \sin \text{ A=0.34566} \\ A=\sin ^{-1}0,34566 \\ A=20.2221 \\ A=20.222^0\text{ (nearest thousandth)} \end{gathered}[/tex]
To calculate angle C
[tex]\begin{gathered} \text{Angle A + Angle B + Angle C = 180 degre}es\text{ ( Sum of angles in a triangle)} \\ 20.222+100_{}+C=180 \\ 120.222+C=180 \\ C=180-120.222 \\ C=59.778^0 \end{gathered}[/tex]
