What are the angle measures of triangle ABC?m∠A = 30°, m∠B = 60°, m∠C = 90°m∠A = 90°, m∠B = 60°, m∠C = 30°m∠A = 60°, m∠B = 90°, m∠C = 30°m∠A = 90°, m∠B = 30°, m∠C = 60°

The correct option is:   m∠A = 90°, m∠B = 60°, m∠C = 30°ExplanationGiven sides of the triangle are....Opposite side of ∠A is:  [tex]a= 24[/tex]Opposite side of ∠B is:  [tex]b=12\sqrt{3}[/tex] and Opposite side of ∠C is:  [tex]c=12[/tex]For finding the angles of triangle ABC, we need to use the Cosine rules. So....[tex]cos(A)= \frac{b^2+c^2-a^2}{2bc}\\ \\ cos(A)=\frac{(12\sqrt{3})^2+(12)^2-(24)^2}{2(12\sqrt{3})(12)} \\ \\ cos(A)=\frac{432+144-576}{288\sqrt{3}}=0\\ \\ A=cos^-^1(0)=90degree\\ \\ \\ cos(B)=\frac{a^2+c^2-b^2}{2ac}\\ \\ cos(B)=\frac{(24)^2+(12)^2-(12\sqrt{3})^2}{2(24)(12)}\\ \\ cos(B)=\frac{576+144-432}{576}=\frac{1}{2}\\ \\ B= cos^-^1(\frac{1}{2})=60 degree\\ \\ \\ cos(C)=\frac{a^2+b^2-c^2}{2ab}\\ \\ cos(C)=\frac{(24)^2+(12\sqrt{3})^2-(12)^2}{2(24)(12\sqrt{3})}[/tex][tex]cos(C)=\frac{576+432-144}{576\sqrt{3}}=\frac{3}{2\sqrt{3}}=\frac{\sqrt{3}}{2}\\ \\ C=cos^-^1(\frac{\sqrt{3}}{2})=30 degree[/tex]So, the measures of ∠A, ∠B and ∠C are  90°, 60° and 30° respectively.