# How do you solve the right triangle given a=15,b=11, c=20?

Oct 23, 2015

The angles of this triangle are: $A \approx 51.93$, $B \approx 28.65$, and $C \approx 99.42$

#### Explanation:

This triangle is not right angled one, because it does not satisfy conditions given in Pytagorean Theorem:

${11}^{2} + {15}^{2} = 121 + 225 = 346$ and ${20}^{2} = 400 \ne 346$.

However you can stil solve this triangle using the Cosine Theorem.

This theorem says that in any triangle following equalities are true:

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

${b}^{2} = {a}^{2} + {c}^{2} - 2 a \mathcal{o} s B$

${a}^{2} = {b}^{2} + {c}^{2} - 2 b \mathcal{o} s A$

where $A$ is the angle opposite to side $a$, $B$ is angle opposite to side $b$, and $C$ is angle opposite to side $c$.

Using this theorem we can calculate:

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

${20}^{2} = {11}^{2} + {15}^{2} - 2 \cdot 11 \cdot 15 \cdot \cos C$

$400 = 121 + 225 - 330 \cos C$

$400 = 346 - 330 \cos C$

$54 = - 330 \cos C$

$\cos C = - \frac{54}{330}$

$C \approx 99.42 \mathrm{de} g$

Using similar calculations you can find, that: $A \approx 51.93$

To calculate $B$ you can either use the Law of Cosines again or use the property of any triangle which says that the sum of its angle is equal to 180 degrees.