(N/A) Data: Draw a circle with centre $A$ and radius $5 \, cm$. Take a point $B$ in its exterior such that $AB = 8 \, cm$.
To construct: Through point $B$,draw two tangents to the circle with centre $A$ and radius $5 \, cm$.
Steps of construction:
$1$. Draw a circle with centre $A$ and radius $5 \, cm$ and mark a point $B$ outside the circle such that $AB = 8 \, cm$.
$2$. Draw the line segment $\overline{AB}$.
$3$. Find the midpoint $M$ of $\overline{AB}$ by constructing its perpendicular bisector.
$4$. With $M$ as the centre and $MA$ as the radius,draw a circle that intersects the original circle at points $X$ and $Y$.
$5$. Draw the rays $\overrightarrow{BX}$ and $\overrightarrow{BY}$.
Thus,$\overleftrightarrow{BX}$ and $\overleftrightarrow{BY}$ are the required tangents.
Calculation: In $\triangle AXB$,$\angle AXB = 90^\circ$ (angle in a semicircle). By Pythagoras theorem,$AB^2 = AX^2 + BX^2$.
$8^2 = 5^2 + BX^2 \implies 64 = 25 + BX^2 \implies BX^2 = 39 \implies BX = \sqrt{39} \approx 6.24 \, cm$.
Therefore,the length of the tangents is approximately $6.24 \, cm$.