How do you find the three angles of the triangle with the given vertices: A(1,0) B(4,6) C(-3,5)?

Mar 25, 2018

color(brown)("The three angles of the triangle are "

color(blue)(hat A = 65.22^@, hat B = 55.31^@, hat C = 59.47^@

Explanation:

Given : $A \left(1 , 0\right) , B \left(4 , 6\right) , C \left(- 3 , 5\right)$

Using distance formula,

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

$a = \sqrt{{\left(4 + 3\right)}^{2} + {\left(6 - 5\right)}^{2}} = \sqrt{50}$

$b = \sqrt{{\left(1 + 3\right)}^{2} + {\left(0 - 5\right)}^{2}} = \sqrt{41}$

$c = \sqrt{{\left(1 - 4\right)}^{2} + {\left(0 - 6\right)}^{2}} = \sqrt{45}$

Applying cosines law,

$\cos A = \frac{b 2 + {c}^{2} - {a}^{2}}{2 b c}$

$\cos A = \frac{41 + 45 - 50}{2 \sqrt{41} \sqrt{45}} \approx 0.4191$

$\hat{A} = {\cos}^{- 1} 0.4191 = {65.22}^{\circ}$

$\cos B = \frac{50 + 45 - 41}{2 \sqrt{45} \sqrt{50}} \approx 0.5692$

$\hat{B} = {\cos}^{- 1} 0.5692 = {55.31}^{\circ}$

$\cos C = \frac{50 + 41 - 45}{2 \sqrt{50} \sqrt{41}} \approx 0.508$

$\hat{C} = {\cos}^{- 1} 0.508 = {59.47}^{\circ}$