Answer :
To find the interior angles of each triangle, we will calculate the angles using the Law of Cosines based on the coordinates of the vertices provided.
### Triangle (a) with vertices [tex]\( A (-4,-5), B (-1,1), C (0,-3) \)[/tex]
1. Calculate the lengths of each side:
[tex]\[ AB = \sqrt{((-1) - (-4))^2 + (1 - (-5))^2} = \sqrt{3^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5} \][/tex]
[tex]\[ BC = \sqrt{(0 - (-1))^2 + ((-3) - 1)^2} = \sqrt{1^2 + (-4)^2} = \sqrt{1 + 16} = \sqrt{17} \][/tex]
[tex]\[ CA = \sqrt{(0 - (-4))^2 + ((-3) - (-5))^2} = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \][/tex]
2. Apply the Law of Cosines to find each angle:
[tex]\[ \cos(A) = \frac{b^2 + c^2 - a^2}{2bc} \Rightarrow \cos(A) = \frac{(3\sqrt{5})^2 + (\sqrt{17})^2 - (2\sqrt{5})^2}{2(3\sqrt{5})(\sqrt{17})} \][/tex]
Therefore, [tex]\( \angle A \approx 36.87^\circ \)[/tex]
[tex]\[ \cos(B) = \frac{c^2 + a^2 - b^2}{2ca} \Rightarrow \cos(B) = \frac{(\sqrt{17})^2 + (2\sqrt{5})^2 - (3\sqrt{5})^2}{2(\sqrt{17})(2\sqrt{5})} \][/tex]
Therefore, [tex]\( \angle B \approx 40.60^\circ \)[/tex]
[tex]\[ \cos(C) = \frac{a^2 + b^2 - c^2}{2ab} \Rightarrow \cos(C) = \frac{(2\sqrt{5})^2 + (3\sqrt{5})^2 - (\sqrt{17})^2}{2(2\sqrt{5})(3\sqrt{5})} \][/tex]
Therefore, [tex]\( \angle C \approx 102.53^\circ \)[/tex]
### Triangle (b) with vertices [tex]\( P (-2,-1), Q (4,-3), R (1,2) \)[/tex]
1. Calculate the lengths of each side:
[tex]\[ PQ = \sqrt{(4 - (-2))^2 + ((-3) - (-1))^2} = \sqrt{6^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} \][/tex]
[tex]\[ QR = \sqrt{(1 - 4)^2 + (2 - (-3))^2} = \sqrt{(-3)^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34} \][/tex]
[tex]\[ RP = \sqrt{(1 - (-2))^2 + (2 - (-1))^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \][/tex]
2. Apply the Law of Cosines to find each angle:
[tex]\[ \cos(P) = \frac{b^2 + c^2 - a^2}{2bc} \Rightarrow \cos(P) = \frac{(2\sqrt{10})^2 + (\sqrt{18})^2 - (\sqrt{34})^2}{2(2\sqrt{10})(3\sqrt{2})} \][/tex]
Therefore, [tex]\( \angle P \approx 63.43^\circ \)[/tex]
[tex]\[ \cos(Q) = \frac{c^2 + a^2 - b^2}{2ca} \Rightarrow \cos(Q) = \frac{(\sqrt{18})^2 + (\sqrt{34})^2 - (2\sqrt{10})^2}{2(3\sqrt{2})(\sqrt{34})} \][/tex]
Therefore, [tex]\( \angle Q \approx 40.60^\circ \)[/tex]
[tex]\[ \cos(R) = \frac{a^2 + b^2 - c^2}{2ab} \Rightarrow \cos(R) = \frac{(\sqrt{34})^2 + (2\sqrt{10})^2 - (\sqrt{18})^2}{2(\sqrt{34})(2\sqrt{10})} \][/tex]
Therefore, [tex]\( \angle R \approx 75.96^\circ \)[/tex]
### Triangle (c) with vertices [tex]\( L (-4,4), M (-3,1), N (6,2) \)[/tex]
1. Calculate the lengths of each side:
[tex]\[ LM = \sqrt{((-3) - (-4))^2 + (1 - 4)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \][/tex]
[tex]\[ MN = \sqrt{(6 - (-3))^2 + (2 - 1)^2} = \sqrt{9^2 + 1^2} = \sqrt{81 + 1} = \sqrt{82} \][/tex]
[tex]\[ NL = \sqrt{(6 - (-4))^2 + (2 - 4)^2} = \sqrt{10^2 + (-2)^2} = \sqrt{100 + 4} = \sqrt{104} \][/tex]
2. Apply the Law of Cosines to find each angle:
[tex]\[ \cos(L) = \frac{b^2 + c^2 - a^2}{2bc} \Rightarrow \cos(L) = \frac{(\sqrt{10})^2 + (\sqrt{82})^2 - (\sqrt{104})^2}{2(\sqrt{10})(\sqrt{82})} \][/tex]
Therefore, [tex]\( \angle L \approx 60.26^\circ \)[/tex]
[tex]\[ \cos(M) = \frac{c^2 + a^2 - b^2}{2ca} \Rightarrow \cos(M) = \frac{(\sqrt{82})^2 + (\sqrt{104})^2 - (\sqrt{10})^2}{2(\sqrt{82})(\sqrt{104})} \][/tex]
Therefore, [tex]\( \angle M \approx 102.09^\circ \)[/tex]
[tex]\[ \cos(N) = \frac{a^2 + b^2 - c^2}{2ab} \Rightarrow \cos(N) = \frac{(\sqrt{104})^2 + (\sqrt{10})^2 - (\sqrt{82})^2}{2(\sqrt{104})(\sqrt{10})} \][/tex]
Therefore, [tex]\( \angle N \approx 17.65^\circ \)[/tex]
### Summary of Angles:
- Triangle (a): [tex]\( \angle A \approx 36.87^\circ, \angle B \approx 40.60^\circ, \angle C \approx 102.53^\circ \)[/tex]
- Triangle (b): [tex]\( \angle P \approx 63.43^\circ, \angle Q \approx 40.60^\circ, \angle R \approx 75.96^\circ \)[/tex]
- Triangle (c): [tex]\( \angle L \approx 60.26^\circ, \angle M \approx 102.09^\circ, \angle N \approx 17.65^\circ \)[/tex]
And that's the detailed calculation for finding the interior angles of the given triangles.
### Triangle (a) with vertices [tex]\( A (-4,-5), B (-1,1), C (0,-3) \)[/tex]
1. Calculate the lengths of each side:
[tex]\[ AB = \sqrt{((-1) - (-4))^2 + (1 - (-5))^2} = \sqrt{3^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5} \][/tex]
[tex]\[ BC = \sqrt{(0 - (-1))^2 + ((-3) - 1)^2} = \sqrt{1^2 + (-4)^2} = \sqrt{1 + 16} = \sqrt{17} \][/tex]
[tex]\[ CA = \sqrt{(0 - (-4))^2 + ((-3) - (-5))^2} = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \][/tex]
2. Apply the Law of Cosines to find each angle:
[tex]\[ \cos(A) = \frac{b^2 + c^2 - a^2}{2bc} \Rightarrow \cos(A) = \frac{(3\sqrt{5})^2 + (\sqrt{17})^2 - (2\sqrt{5})^2}{2(3\sqrt{5})(\sqrt{17})} \][/tex]
Therefore, [tex]\( \angle A \approx 36.87^\circ \)[/tex]
[tex]\[ \cos(B) = \frac{c^2 + a^2 - b^2}{2ca} \Rightarrow \cos(B) = \frac{(\sqrt{17})^2 + (2\sqrt{5})^2 - (3\sqrt{5})^2}{2(\sqrt{17})(2\sqrt{5})} \][/tex]
Therefore, [tex]\( \angle B \approx 40.60^\circ \)[/tex]
[tex]\[ \cos(C) = \frac{a^2 + b^2 - c^2}{2ab} \Rightarrow \cos(C) = \frac{(2\sqrt{5})^2 + (3\sqrt{5})^2 - (\sqrt{17})^2}{2(2\sqrt{5})(3\sqrt{5})} \][/tex]
Therefore, [tex]\( \angle C \approx 102.53^\circ \)[/tex]
### Triangle (b) with vertices [tex]\( P (-2,-1), Q (4,-3), R (1,2) \)[/tex]
1. Calculate the lengths of each side:
[tex]\[ PQ = \sqrt{(4 - (-2))^2 + ((-3) - (-1))^2} = \sqrt{6^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} \][/tex]
[tex]\[ QR = \sqrt{(1 - 4)^2 + (2 - (-3))^2} = \sqrt{(-3)^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34} \][/tex]
[tex]\[ RP = \sqrt{(1 - (-2))^2 + (2 - (-1))^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \][/tex]
2. Apply the Law of Cosines to find each angle:
[tex]\[ \cos(P) = \frac{b^2 + c^2 - a^2}{2bc} \Rightarrow \cos(P) = \frac{(2\sqrt{10})^2 + (\sqrt{18})^2 - (\sqrt{34})^2}{2(2\sqrt{10})(3\sqrt{2})} \][/tex]
Therefore, [tex]\( \angle P \approx 63.43^\circ \)[/tex]
[tex]\[ \cos(Q) = \frac{c^2 + a^2 - b^2}{2ca} \Rightarrow \cos(Q) = \frac{(\sqrt{18})^2 + (\sqrt{34})^2 - (2\sqrt{10})^2}{2(3\sqrt{2})(\sqrt{34})} \][/tex]
Therefore, [tex]\( \angle Q \approx 40.60^\circ \)[/tex]
[tex]\[ \cos(R) = \frac{a^2 + b^2 - c^2}{2ab} \Rightarrow \cos(R) = \frac{(\sqrt{34})^2 + (2\sqrt{10})^2 - (\sqrt{18})^2}{2(\sqrt{34})(2\sqrt{10})} \][/tex]
Therefore, [tex]\( \angle R \approx 75.96^\circ \)[/tex]
### Triangle (c) with vertices [tex]\( L (-4,4), M (-3,1), N (6,2) \)[/tex]
1. Calculate the lengths of each side:
[tex]\[ LM = \sqrt{((-3) - (-4))^2 + (1 - 4)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \][/tex]
[tex]\[ MN = \sqrt{(6 - (-3))^2 + (2 - 1)^2} = \sqrt{9^2 + 1^2} = \sqrt{81 + 1} = \sqrt{82} \][/tex]
[tex]\[ NL = \sqrt{(6 - (-4))^2 + (2 - 4)^2} = \sqrt{10^2 + (-2)^2} = \sqrt{100 + 4} = \sqrt{104} \][/tex]
2. Apply the Law of Cosines to find each angle:
[tex]\[ \cos(L) = \frac{b^2 + c^2 - a^2}{2bc} \Rightarrow \cos(L) = \frac{(\sqrt{10})^2 + (\sqrt{82})^2 - (\sqrt{104})^2}{2(\sqrt{10})(\sqrt{82})} \][/tex]
Therefore, [tex]\( \angle L \approx 60.26^\circ \)[/tex]
[tex]\[ \cos(M) = \frac{c^2 + a^2 - b^2}{2ca} \Rightarrow \cos(M) = \frac{(\sqrt{82})^2 + (\sqrt{104})^2 - (\sqrt{10})^2}{2(\sqrt{82})(\sqrt{104})} \][/tex]
Therefore, [tex]\( \angle M \approx 102.09^\circ \)[/tex]
[tex]\[ \cos(N) = \frac{a^2 + b^2 - c^2}{2ab} \Rightarrow \cos(N) = \frac{(\sqrt{104})^2 + (\sqrt{10})^2 - (\sqrt{82})^2}{2(\sqrt{104})(\sqrt{10})} \][/tex]
Therefore, [tex]\( \angle N \approx 17.65^\circ \)[/tex]
### Summary of Angles:
- Triangle (a): [tex]\( \angle A \approx 36.87^\circ, \angle B \approx 40.60^\circ, \angle C \approx 102.53^\circ \)[/tex]
- Triangle (b): [tex]\( \angle P \approx 63.43^\circ, \angle Q \approx 40.60^\circ, \angle R \approx 75.96^\circ \)[/tex]
- Triangle (c): [tex]\( \angle L \approx 60.26^\circ, \angle M \approx 102.09^\circ, \angle N \approx 17.65^\circ \)[/tex]
And that's the detailed calculation for finding the interior angles of the given triangles.
