Answer:
[tex]A = 74.7^\circ[/tex]
[tex]B = 42.5^\circ[/tex]
[tex]C = 62.8^\circ[/tex]
Step-by-step explanation:
Given
[tex]A = (-1,2) \to (x_1,y_1)[/tex]
[tex]B = (2,8) \to (x_2,y_2)[/tex]
[tex]C = (4,1) \to (x_3,y_3)[/tex]
Required
The measure of each angle
First, we calculate the length of the three sides of the triangle.
This is calculated using distance formula
[tex]d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2[/tex]
For AB
[tex]A = (-1,2) \to (x_1,y_1)[/tex]
[tex]B = (2,8) \to (x_2,y_2)[/tex]
[tex]d = \sqrt{(-1 - 2)^2 + (2 - 8)^2[/tex]
[tex]d = \sqrt{(-3)^2 + (-6)^2[/tex]
[tex]d = \sqrt{45[/tex]
So:
[tex]AB = \sqrt{45[/tex]
For BC
[tex]B = (2,8) \to (x_2,y_2)[/tex]
[tex]C = (4,1) \to (x_3,y_3)[/tex]
[tex]BC = \sqrt{(2 - 4)^2 + (8 - 1)^2[/tex]
[tex]BC = \sqrt{(-2)^2 + (7)^2[/tex]
[tex]BC = \sqrt{53[/tex]
For AC
[tex]A = (-1,2) \to (x_1,y_1)[/tex]
[tex]C = (4,1) \to (x_3,y_3)[/tex]
[tex]AC = \sqrt{(-1 - 4)^2 + (2 - 1)^2[/tex]
[tex]AC = \sqrt{(-5)^2 + (1)^2[/tex]
[tex]AC = \sqrt{26[/tex]
So, we have:
[tex]AB = \sqrt{45[/tex]
[tex]BC = \sqrt{53[/tex]
[tex]AC = \sqrt{26[/tex]
By representation
[tex]AB \to c[/tex]
[tex]BC \to a[/tex]
[tex]AC \to b[/tex]
So, we have:
[tex]a = \sqrt{53[/tex]
[tex]b = \sqrt{26[/tex]
[tex]c = \sqrt{45[/tex]
By cosine laws, the angles are calculated using:
[tex]a^2 = b^2 + c^2 -2bc \cos A[/tex]
[tex]b^2 = a^2 + c^2 -2ac \cos B[/tex]
[tex]c^2 = a^2 + b^2 -2ab\ cos C[/tex]
[tex]a^2 = b^2 + c^2 -2bc \cos A[/tex]
[tex](\sqrt{53})^2 = (\sqrt{26})^2 +(\sqrt{45})^2 - 2 * (\sqrt{26}) +(\sqrt{45}) * \cos A[/tex]
[tex]53 = 26 +45 - 2 * 34.21 * \cos A[/tex]
[tex]53 = 26 +45 - 68.42 * \cos A[/tex]
Collect like terms
[tex]53 - 26 -45 = - 68.42 * \cos A[/tex]
[tex]-18 = - 68.42 * \cos A[/tex]
Solve for [tex]\cos A[/tex]
[tex]\cos A =\frac{-18}{-68.42}[/tex]
[tex]\cos A =0.2631[/tex]
Take arc cos of both sides
[tex]A =\cos^{-1}(0.2631)[/tex]
[tex]A = 74.7^\circ[/tex]
[tex]b^2 = a^2 + c^2 -2ac \cos B[/tex]
[tex](\sqrt{26})^2 = (\sqrt{53})^2 +(\sqrt{45})^2 - 2 * (\sqrt{53}) +(\sqrt{45}) * \cos B[/tex]
[tex]26 = 53 +45 -97.67 * \cos B[/tex]
Collect like terms
[tex]26 - 53 -45= -97.67 * \cos B[/tex]
[tex]-72= -97.67 * \cos B[/tex]
Solve for [tex]\cos B[/tex]
[tex]\cos B = \frac{-72}{-97.67}[/tex]
[tex]\cos B = 0.7372[/tex]
Take arc cos of both sides
[tex]B = \cos^{-1}(0.7372)[/tex]
[tex]B = 42.5^\circ[/tex]
For the third angle, we use:
[tex]A + B + C = 180[/tex] --- angles in a triangle
Make C the subject
[tex]C = 180 - A -B[/tex]
[tex]C = 180 - 74.7 -42.5[/tex]
[tex]C = 62.8^\circ[/tex]
