Angle ratios

Since the largest angle subtends the largest side, we have $\angle ABC>\angle BAC>\angle BCA$ , so we are trying to find the value of $a$ , where $\left | \angle ABC \right |=a\left | \angle BCA \right |$ .

This is because the largest angle of a triangle is always opposite the longest side, and the smallest angle is always opposite the smallest side.

We conjecture $a=2$ , and so must prove $\left | \angle ABC \right |=2\left | \angle BCA \right |$ .

Let the angle bisector of $\angle ABC$ meet $AC$ at $D$ . Then by the Angle bisector theorem

$\dfrac{AD}{DC}=\dfrac{AB}{BC}=\dfrac{4}{5}$ .

We also know that $AD+DC=6$ , so

$\dfrac{4}{5}DC+DC=6\Longrightarrow DC=\dfrac{30}{9}$ and $AD=\dfrac{24}{9}$ .

Now $\dfrac{AD}{AB}=\dfrac{24}{9\times 4}=\dfrac{2}{3}=\dfrac{AB}{AC}$ , and since $\angle BAD=\angle BAC$ , we know that

$\bigtriangleup BAD \sim \bigtriangleup CAB\Longrightarrow \angle ABD=\angle BCA$ .

Finally, since $\angle ABD=\dfrac{1}{2} \angle ABC$ , we know that $\angle ABC=2\angle BCA$ .

Hence, $a=\boxed{2}$ .

The largest angle in a triangle is always opposite the longest side while the smallest angle is always opposite the shortest side. From the figure, $\angle B$ is the largest while $\angle C$ is the smallest.

By Cosine Law, we have

$6^2=4^2+5^2-2(4)(5)cos~B$ $\color{#D61F06}\implies$ $\boxed{\angle B=82.81924422^\circ}$

$4^2=5^2+6^2-2(5)(6)cos~C$ $\color{#D61F06}\implies$ $\boxed{\angle C=41.40962211^\circ}$

It follows that,

$\dfrac{\angle B}{\angle C}=2$

$\large\therefore$ $\angle B$ is two times larger than $\angle C$