What is the distance between the points

We have to find the distance between $A\left(5 \sin 60^{\circ}, 0\right)$ and $B\left(0,5 \sin 30^{\circ}\right)$.

In general, the distance between $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ is given by,

$\mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

So,

$\mathrm{AB}=\sqrt{\left(5 \sin 60^{\circ}-0\right)^{2}+\left(0-5 \sin 30^{\circ}\right)^{2}}$

But according to the trigonometric identity,

$\sin ^{2} \theta+\cos ^{2} \theta=1$

And,

$\sin 30^{\circ}=\cos 60^{\circ}$

Therefore,

$\mathrm{AB}=\sqrt{5^{2}\left(\sin ^{2} 60^{\circ}+\cos ^{2} 60^{\circ}\right)}$

$=5$