# Triangle ABC is formed by the points A(3, 4), B(2, 6), and C(5, 8). Find angle A to the nearest tenth of a degree?

## Triangle ABC is formed by the points A(3, 4), B(2, 6), and C(5, 8). Find angle A to the nearest tenth of a degree.

Apr 11, 2018

The measure of angle $A \approx {53.1}^{\circ}$

#### Explanation:

I will solve this using the Law of Cosines which states that for a triangle, with side lengths, $a$, $b$, and $c$, and angle $A$ opposite the side with length $a$

${a}^{2} = {b}^{2} + {c}^{2} - 2 \cdot b \cdot c \cdot \cos \left(A\right)$

Solving this for $\cos A$ we have

$A = {\cos}^{-} 1 \left(\frac{{b}^{2} + {c}^{2} - {a}^{2}}{2 b c}\right)$ Let's first calculate $a$, $b$, and $c$ using the distance formula.

$a = \sqrt{{\left(8 - 6\right)}^{2} + {\left(5 - 2\right)}^{2}} = \sqrt{13}$

$b = \sqrt{{\left(8 - 4\right)}^{2} + {\left(5 - 3\right)}^{2}} = 2 \sqrt{5}$

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

Plugging these values into our formula for A gives

$A = {\cos}^{-} 1 \left[\frac{{\left(2 \sqrt{5}\right)}^{2} + {\left(\sqrt{5}\right)}^{2} - {\left(\sqrt{13}\right)}^{2}}{2 \cdot 2 \sqrt{5} \sqrt{5}}\right]$

$= {\cos}^{-} 1 \left(\frac{20 + 5 - 13}{20}\right) = {\cos}^{-} 1 \left(\frac{3}{5}\right) \approx {53.1}^{\circ}$