Question

Item 7

△ABC has vertices A(−1, 2), B(2, 8), and C(4, 1). Find the measure of each angle of the triangle. Round decimal answers to the nearest tenth.

Answer 1

`A = 74.7^\circ`

`B = 42.5^\circ`

`C = 62.8^\circ`

Explanation:

Given

`A = (-1,2) \to (x_1,y_1)`

`B = (2,8) \to (x_2,y_2)`

`C = (4,1) \to (x_3,y_3)`

Required

The measure of each angle

First, we calculate the length of the three sides of the triangle.

This is calculated using distance formula

`d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2`

For AB

`A = (-1,2) \to (x_1,y_1)`

`B = (2,8) \to (x_2,y_2)`

`d = \sqrt{(-1 - 2)^2 + (2 - 8)^2`

`d = \sqrt{(-3)^2 + (-6)^2`

`d = \sqrt{45`

So:

`AB = \sqrt{45`

For BC

`B = (2,8) \to (x_2,y_2)`

`C = (4,1) \to (x_3,y_3)`

`BC = \sqrt{(2 - 4)^2 + (8 - 1)^2`

`BC = \sqrt{(-2)^2 + (7)^2`

`BC = \sqrt{53`

For AC

`A = (-1,2) \to (x_1,y_1)`

`C = (4,1) \to (x_3,y_3)`

`AC = \sqrt{(-1 - 4)^2 + (2 - 1)^2`

`AC = \sqrt{(-5)^2 + (1)^2`

`AC = \sqrt{26`

So, we have:

`AB = \sqrt{45`

`BC = \sqrt{53`

`AC = \sqrt{26`

By representation

`AB \to c`

`BC \to a`

`AC \to b`

So, we have:

`a = \sqrt{53`

`b = \sqrt{26`

`c = \sqrt{45`

By cosine laws, the angles are calculated using:

`a^2 = b^2 + c^2 -2bc \cos A`

`b^2 = a^2 + c^2 -2ac \cos B`

`c^2 = a^2 + b^2 -2ab\ cos C`

`a^2 = b^2 + c^2 -2bc \cos A`

`(√(53))^2 = (√(26))^2 +(√(45))^2 - 2 * (√(26)) +(√(45)) * \cos A`

`53 = 26 +45 - 2 * 34.21 * \cos A`

`53 = 26 +45 - 68.42 * \cos A`

Collect like terms

`53 - 26 -45 = - 68.42 * \cos A`

`-18 = - 68.42 * \cos A`

Solve for
`\cos A`

`\cos A =(-18)/(-68.42)`

`\cos A =0.2631`

Take arc cos of both sides

`A =\cos^(-1)(0.2631)`

`A = 74.7^\circ`

`b^2 = a^2 + c^2 -2ac \cos B`

`(√(26))^2 = (√(53))^2 +(√(45))^2 - 2 * (√(53)) +(√(45)) * \cos B`

`26 = 53 +45 -97.67 * \cos B`

Collect like terms

`26 - 53 -45= -97.67 * \cos B`

`-72= -97.67 * \cos B`

Solve for
`\cos B`

`\cos B = (-72)/(-97.67)`

`\cos B = 0.7372`

Take arc cos of both sides

`B = \cos^(-1)(0.7372)`

`B = 42.5^\circ`

For the third angle, we use:

`A + B + C = 180` --- angles in a triangle

Make C the subject

`C = 180 - A -B`

`C = 180 - 74.7 -42.5`

`C = 62.8^\circ`