Answer:
a bit more than 48°
Step-by-step explanation:
We use the law of cosines for this.
The theorem is an extension of pythagoras' theorem, working for all triangles. In our case
[tex]\overline{BC}^2=\overline{AB}^2+\overline{AC}^2-2\overline{AB}\cdot\overline{AC}cos \hat A[/tex]
Let's replace all the values we have and find whatever is left.
[tex]10^2=11^2+13^2-2(11)(13)cos \hat A\\-286cos\hat A = -190\\cos \hat A\approx 0.6643 \rightarrow \hat A \approx 48\°[/tex]
