Step 1
Given;
Step 2
[tex]A)\text{ Determine m}\angle A[/tex]
Using cosine rule;
[tex]\begin{gathered} a^2=b^2+c^2-2bccosA \\ 16^2=(16\sqrt{3})^3+32^2-2(32)(16\sqrt{3})cosA \end{gathered}[/tex][tex]256=768+1024-1773.620037cos\text{ A}[/tex][tex]\begin{gathered} -1536=-1773.620037cos\text{ A} \\ A=30.00000056^o \end{gathered}[/tex]
B) Below is a unit circle image
[tex]\begin{gathered} At\text{ a point in our solution we had;} \\ -1536=-2(32)(16\sqrt{3})cosA \\ cosA=\frac{\sqrt{3}}{2} \end{gathered}[/tex]
From the unit circle, we search for Tan A in the first quadrant since cos A is in the first quadrant.
[tex]\begin{gathered} TanA=\frac{y}{x} \\ y=\frac{1}{2},x=\frac{\sqrt{3}}{2} \\ TanA=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} \\ TanA=\frac{1}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3} \\ TanA=\frac{\sqrt{3}}{3} \end{gathered}[/tex]
C) Calculate the area of triangle ABC
Using
[tex]\begin{gathered} Area=\frac{1}{2}bcsinA \\ Area=\frac{1}{2}\times16\sqrt{3}\times32\times sin(30) \\ Area=221.7025034in^2 \end{gathered}[/tex]
Answer;
[tex]\begin{gathered} A)m\angle A=30^o \\ B)TanA\text{ =}\frac{\sqrt{3}}{3} \\ C)AreaoftriangleABC=221.7025034in^2 \end{gathered}[/tex]
