If $$| \overrightarrow a |=13 | \overrightarrow b |=5\: and \: \overrightarrow a.\overrightarrow b=60$$ then find $$|\overrightarrow a$$ x $$\overrightarrow b|$$

Toolbox:
• $\cos\theta=\large\frac{\overrightarrow a.\overrightarrow b}{\mid\overrightarrow a\mid\mid\overrightarrow b\mid}$
• $\mid \overrightarrow a\times\overrightarrow b\mid=\mid\overrightarrow a\mid\mid\overrightarrow b\mid.\sin\theta$
Step 1:
Given :$\mid\overrightarrow a\mid=13$ and $\mid\overrightarrow b\mid$ and $\overrightarrow a.\overrightarrow b=60$
$\overrightarrow a.\overrightarrow b=\mid\overrightarrow a\mid\mid\overrightarrow b\mid\cos\theta$
$\cos\theta=\large\frac{\overrightarrow a.\overrightarrow b}{\mid\overrightarrow a\mid\mid\overrightarrow b\mid}$
$\qquad=\large\frac{60}{13\times 5}$
$\cos\theta=\large\frac{12}{13}$
Step 2:
We can find the third side of the triangle by using Pythagoras theorem.
$x^2+12^2=13^2$
$x^2=25$
$x=5$
$\sin\theta=\large\frac{5}{13}$
Step 3:
$\overrightarrow a\times\overrightarrow b=\mid\overrightarrow a \mid\mid\overrightarrow b\mid\sin\theta .\hat n$
$\mid\overrightarrow a\times\overrightarrow b\mid=\mid\overrightarrow a \mid\mid\overrightarrow b\mid\sin\theta$
$\Rightarrow 13\times 5\times \large\frac{5}{13}$
$\Rightarrow 25$