In a rectangle ABCD, AB = 20cm, $\angle BAC=60$ , calculate the side BC and the diagonals AC and BD.

In the above diagram AB = 20cm.
Now it is given $\angle A$ = 60,
Now in the triangle ABC,
We know that the formula for cos and tan are,
$\begin{align}
  & \cos A=\dfrac{height}{hypotenuse} \\
 & \tan A=\dfrac{base}{height} \\
\end{align}$
The diagonals are AC and BD, we will take the triangle ABC to use the trigonometric formula to find the value of AC and BC.
Now substituting the values of A = 60, height = AB = 20, hypotenuse = AC, and base = BC we get,
We know that $\tan 60=\sqrt{3}$ ,
$\begin{align}
  & \tan 60=\dfrac{BC}{AB} \\
 & BC=20\sqrt{3} \\
\end{align}$
Now in the formula of cos we get,
We know that $\cos 60=\dfrac{1}{2}$ ,
$\begin{align}
  & \cos A=\dfrac{AB}{AC} \\
 & AC=\dfrac{20}{\cos 60} \\
 & AC=40 \\
\end{align}$
Now we know that in a rectangle the length of diagonals are equal so AC will be equal to BD.
Hence, BD = AC = 40cm and BC = $20\sqrt{3}cm$ .
Hence, we have all the values that have been asked in the question.

Note: To solve this question one can also take the other triangle ACD, and then we will have to substitute the values in these two formulas to find the value of the other two sides of the triangle. One can also use Pythagoras theorem ${{b}^{2}}={{c}^{2}}-{{a}^{2}}$ to find the value of third side given that the other two sides are known.