# Trianlge A B C has A B = 33 , A C = 88 , B C =

Trianlge $ABC$ has $AB=33,AC=88,BC=77.$ Point $D$ lies on $BC$ with $BD=21.$ Compute $\mathrm{\angle }BAD.$
By the angle bisector theorem we have $\frac{21}{56}=\frac{33}{88}$ and we know that $AD$ bisects $\mathrm{\angle }BAC$. Now from here I was able to use the law of cosines, but it resulted in a pretty ugly looking expression
${77}^{2}={33}^{2}+{88}^{2}-2\left(33\right)\left(88\right)cos\left(\mathrm{\angle }BAC\right)$
from here I was pretty much forced to use a calculator to figure out that
$\mathrm{\angle }BAC=60⇒\mathrm{\angle }BAD=30.$
After looking at the solution they had a very similar approach, but instead of
${77}^{2}={33}^{2}+{88}^{2}-2\left(33\right)\left(88\right)cos\left(\mathrm{\angle }BAC\right)$
${7}^{2}={3}^{2}+{8}^{2}-2\left(3\right)\left(8\right)cos\left(\mathrm{\angle }BAC\right)$
which I don’t really get. How can you take one digit of from every term there?
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Oliver Shepherd
This isn't a question of geometry.....its simple arithmetic:
${77}^{2}={33}^{2}+{88}^{2}-2\left(33\right)\left(88\right)cos\left(\mathrm{\angle }BAC\right)$
Here you can see every term has a factor of ${11}^{2}$.
$\left({11}^{2}\right){7}^{2}={3}^{2}\left(11{\right)}^{2}+{8}^{2}\left(11{\right)}^{2}-\left({11}^{2}\right)2\left(3\right)\left(8\right)cos\left(\mathrm{\angle }BAC\right)$
You can cancel out this factor by dividing by ${11}^{2}$ on both sides.
${7}^{2}={3}^{2}+{8}^{2}-2\left(3\right)\left(8\right)cos\left(\mathrm{\angle }BAC\right)$