[tex]\begin{gathered} c=7.6 \\ \angle A=37 \end{gathered}[/tex]
Explanation
to solve this we need to use the cosine law
Law of cosines says
[tex]\begin{gathered} a^2=b^2+c^2-2bc\cdot\cos (A) \\ b^2=a^2+c^2-2ac\cdot\cos (B) \\ c^2=a^2+b^2-2ab\cdot\cos (C) \end{gathered}[/tex]
then
Step 1
find c
Let
[tex]\begin{gathered} b=8 \\ a=5 \\ \angle C=67 \end{gathered}[/tex]
now, let's find c
[tex]\begin{gathered} c^2=a^2+b^2-2bc\cdot\cos (C) \\ \text{replace} \\ c^2=5^2+8^2-2\cdot5\cdot8\cdot\cos (67) \\ c^2=25+64-31.25849028 \\ c^2=57.74 \\ \text{square root in both sides} \\ \sqrt{c^2}=\sqrt{57.74} \\ c=7.598 \\ \text{rounded} \\ c=7.6 \end{gathered}[/tex]
Step 2
now,let's find the angle A
[tex]\begin{gathered} a^2=b^2+c^2-2bc\cdot\cos (A) \\ replace \\ 5^2=8^2+7.6^2-2\cdot8\cdot7.6\cdot\cos (A) \\ 25=64+57.76-121.6\text{ cos(A)} \\ 25=121.76-121.6(\cos A) \\ \text{subtract 121.76 in both sides} \\ 25-121.76=121.76-121.6(\cos A)-121.76 \\ -96.76=-121.6(\cos A) \\ \text{divide both sides by -121.6} \\ \frac{-96.76}{-121.6}=\frac{-121.6(\cos A)}{-121.6} \\ 0.795=\cos \text{ (A)} \\ Inverse\text{ cosine} \\ \cos ^{-1}(0.795)=\cos ^{-1}(\cos \text{ (A))} \\ 37.276=\text{ A} \\ \text{rounded} \\ A=37\~\text{ =A} \end{gathered}[/tex]
Step 3
Finally,let's find angle B
we can use the formula as we did in step 2 to find angle A, but also we can use this fact:
the sum of the internal angles in a triangle equals 180 ,so
[tex]\angle A+\angle B+\angle C=180[/tex]
replace and solve for angle C
[tex]\begin{gathered} \angle A+\angle B+\angle C=180 \\ 37+\angle B+67=180 \\ 104+\angle B=180 \\ \text{subtract 104 in both sides} \\ 104+\angle B-104=180-104 \\ \angle B=76 \end{gathered}[/tex]
