Solve the triangle whose vertices are $A(4,2), B(6,11)$, and $C(7,3)$.
Answer
Final Answer: The lengths of the sides of the triangle are approximately \(AB = 9.22\), \(BC = 8.06\), and \(CA = 3.16\). The measures of the angles are approximately \(A = 101.31^\circ\), \(B = 59.04^\circ\), and \(C = 19.65^\circ\). The area of the triangle is approximately \(12.50\) square units. Therefore, the solution to the triangle is \(\boxed{AB = 9.22, BC = 8.06, CA = 3.16, A = 101.31^\circ, B = 59.04^\circ, C = 19.65^\circ, \text{Area} = 12.50}\).
Step 1 ：First, we find the lengths of the sides of the triangle using the distance formula. The distance between points A and B is \(\sqrt{(6-4)^2 + (11-2)^2} = 9.22\). The distance between points B and C is \(\sqrt{(7-6)^2 + (3-11)^2} = 8.06\). The distance between points C and A is \(\sqrt{(4-7)^2 + (2-3)^2} = 3.16\).
Step 2 ：Next, we find the measures of the angles using the law of cosines. The measure of angle A is \(\cos^{-1}\left(\frac{8.06^2 + 3.16^2 - 9.22^2}{2 \cdot 8.06 \cdot 3.16}\right) = 101.31^\circ\). The measure of angle B is \(\cos^{-1}\left(\frac{9.22^2 + 3.16^2 - 8.06^2}{2 \cdot 9.22 \cdot 3.16}\right) = 59.04^\circ\). The measure of angle C is \(\cos^{-1}\left(\frac{9.22^2 + 8.06^2 - 3.16^2}{2 \cdot 9.22 \cdot 8.06}\right) = 19.65^\circ\).
Step 3 ：Finally, we find the area of the triangle using Heron's formula. The semi-perimeter of the triangle is \(\frac{9.22 + 8.06 + 3.16}{2} = 10.22\). The area of the triangle is \(\sqrt{10.22(10.22-9.22)(10.22-8.06)(10.22-3.16)} = 12.50\) square units.
Step 4 ：Final Answer: The lengths of the sides of the triangle are approximately \(AB = 9.22\), \(BC = 8.06\), and \(CA = 3.16\). The measures of the angles are approximately \(A = 101.31^\circ\), \(B = 59.04^\circ\), and \(C = 19.65^\circ\). The area of the triangle is approximately \(12.50\) square units. Therefore, the solution to the triangle is \(\boxed{AB = 9.22, BC = 8.06, CA = 3.16, A = 101.31^\circ, B = 59.04^\circ, C = 19.65^\circ, \text{Area} = 12.50}\).
