# How do you find angles A, B, and C if in triangle ABC, a=12, b=15, and c=20?

May 3, 2018

$A \approx {36.71}^{\circ}$
$B \approx {48.34}^{\circ}$
$C \approx {94.93}^{\circ}$

#### Explanation:

To solve the for the angles when we have the lengths of all three sides, we use the law of cosines.

The law of cosines states that angle $A = {\cos}^{-} 1 \left(\frac{{a}^{2} - {b}^{2} - {c}^{2}}{- 2 b c}\right)$, $B = {\cos}^{-} 1 \left(\frac{{b}^{2} - {a}^{2} - {c}^{2}}{- 2 a c}\right)$, and $C = {\cos}^{-} 1 \left(\frac{{c}^{2} - {a}^{2} - {b}^{2}}{- 2 a b}\right)$.

Let's find angle $A$ first:
$A = \frac{{a}^{2} - {b}^{2} - {c}^{2}}{- 2 b c}$

$A = \frac{{12}^{2} - {15}^{2} - {20}^{2}}{- 2 \left(15\right) \left(20\right)}$

$A = {\cos}^{-} 1 \left(\frac{- 481}{- 600}\right)$

$A \approx {36.71}^{\circ}$

Now angle $B$:
$B = {\cos}^{-} 1 \left(\frac{{b}^{2} - {a}^{2} - {c}^{2}}{- 2 a c}\right)$

$B = {\cos}^{-} 1 \left(\frac{{15}^{2} - {12}^{2} - {20}^{2}}{- 2 \left(12\right) \left(20\right)}\right)$

$B = {\cos}^{-} 1 \left(\frac{- 319}{- 480}\right)$

$B \approx {48.34}^{\circ}$

Finally angle $C$:
$C = {\cos}^{-} 1 \left(\frac{{c}^{2} - {a}^{2} - {b}^{2}}{- 2 a b}\right)$

$C = {\cos}^{-} 1 \left(\frac{{20}^{2} - {12}^{2} - {15}^{2}}{- 2 \left(12\right) \left(15\right)}\right)$

$C = {\cos}^{-} 1 \left(\frac{31}{- 360}\right)$

$C \approx {94.93}^{\circ}$

We can also find angle $C$ by doing ${180}^{\circ} - {36.71}^{\circ} - {48.34}^{\circ}$, since the measures of the angles in a triangle add up to ${180}^{\circ}$.

Hope this helps!