A triangle is defined by the three points: A=(7,9) B=(8,8) C=(6,6). How to determine all three angles in the triangle (in radians) ?

$A = {63.435}^{\setminus} \circ$, $B = {90}^{\setminus} \circ$, $C = {26.565}^{\setminus} \circ$

Explanation:

The lengths of sides of triangle having vertices $A \left(7 , 9\right)$, $B \left(8 , 8\right)$ & $C \left(6 , 6\right)$ are given as

$A B = \setminus \sqrt{{\left(7 - 8\right)}^{2} + {\left(9 - 8\right)}^{2}} = \setminus \sqrt{2}$

$B C = \setminus \sqrt{{\left(8 - 6\right)}^{2} + {\left(8 - 6\right)}^{2}} = 2 \setminus \sqrt{2}$

$A C = \setminus \sqrt{{\left(7 - 6\right)}^{2} + {\left(9 - 6\right)}^{2}} = \setminus \sqrt{10}$

Now, using Cosine rule in $\setminus \Delta A B C$ as follows

$\setminus \cos A = \setminus \frac{A {B}^{2} + A {C}^{2} - B {C}^{2}}{2 \left(A B\right) \left(A C\right)}$

$\setminus \cos A = \setminus \frac{{\left(\setminus \sqrt{2}\right)}^{2} + {\left(\setminus \sqrt{10}\right)}^{2} - {\left(2 \setminus \sqrt{2}\right)}^{2}}{2 \left(\setminus \sqrt{2}\right) \left(\setminus \sqrt{10}\right)}$

$\setminus \cos A = \frac{1}{\setminus} \sqrt{5}$

$A = \setminus {\cos}^{- 1} \left(\frac{1}{\setminus} \sqrt{5}\right) = {63.435}^{\setminus} \circ$

Similarly, we get

$\setminus \cos B = \setminus \frac{A {B}^{2} + B {C}^{2} - A {C}^{2}}{2 \left(A B\right) \left(B C\right)}$

$\setminus \cos B = \setminus \frac{{\left(\setminus \sqrt{2}\right)}^{2} + {\left(2 \setminus \sqrt{2}\right)}^{2} - {\left(\setminus \sqrt{10}\right)}^{2}}{2 \left(\setminus \sqrt{2}\right) \left(2 \setminus \sqrt{2}\right)}$

$\setminus \cos B = 0$

$B = {90}^{\setminus} \circ$

$\setminus \cos C = \setminus \frac{A {C}^{2} + B {C}^{2} - A {B}^{2}}{2 \left(A C\right) \left(B C\right)}$

$\setminus \cos C = \setminus \frac{{\left(\setminus \sqrt{10}\right)}^{2} + {\left(2 \setminus \sqrt{2}\right)}^{2} - {\left(\setminus \sqrt{2}\right)}^{2}}{2 \left(\setminus \sqrt{10}\right) \left(2 \setminus \sqrt{2}\right)}$

$\setminus \cos C = \frac{2}{\setminus} \sqrt{5}$

$C = \setminus {\cos}^{- 1} \left(\frac{2}{\setminus} \sqrt{5}\right) = {26.565}^{\setminus} \circ$