Step 1: Given condition
\[
\tan 2A = 1
\]
Step 2: Recall standard tangent values
\[
\tan 45^\circ = 1
\]
Step 3: Equating angles
\[
2A = 45^\circ
\]
Step 4: Solve for $A$
\[
A = \frac{45^\circ}{2} = 22.5^\circ = 22\dfrac{1}{2}^\circ
\]
\[
\boxed{A = 22\dfrac{1}{2}^\circ}
\]
The product of $\sqrt{2}$ and $(2-\sqrt{2})$ will be:
If a tangent $PQ$ at a point $P$ of a circle of radius $5 \,\text{cm}$ meets a line through the centre $O$ at a point $Q$ so that $OQ = 12 \,\text{cm}$, then length of $PQ$ will be:
In the figure $DE \parallel BC$. If $AD = 3\,\text{cm}$, $DE = 4\,\text{cm}$ and $DB = 1.5\,\text{cm}$, then the measure of $BC$ will be: