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

${77}^{2}={33}^{2}+{88}^{2}-2(33)(88)cos(\mathrm{\angle}BAC)$

from here I was pretty much forced to use a calculator to figure out that

$\mathrm{\angle}BAC=60\Rightarrow \mathrm{\angle}BAD=30.$

After looking at the solution they had a very similar approach, but instead of

${77}^{2}={33}^{2}+{88}^{2}-2(33)(88)cos(\mathrm{\angle}BAC)$

they had

${7}^{2}={3}^{2}+{8}^{2}-2(3)(8)cos(\mathrm{\angle}BAC)$

which I don’t really get. How can you take one digit of from every term there?

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

${77}^{2}={33}^{2}+{88}^{2}-2(33)(88)cos(\mathrm{\angle}BAC)$

from here I was pretty much forced to use a calculator to figure out that

$\mathrm{\angle}BAC=60\Rightarrow \mathrm{\angle}BAD=30.$

After looking at the solution they had a very similar approach, but instead of

${77}^{2}={33}^{2}+{88}^{2}-2(33)(88)cos(\mathrm{\angle}BAC)$

they had

${7}^{2}={3}^{2}+{8}^{2}-2(3)(8)cos(\mathrm{\angle}BAC)$

which I don’t really get. How can you take one digit of from every term there?