Angles and Polygons: Question 3

Worked solution

Step 1: Find the interior angle of the regular pentagon

A pentagon has $n = 5$ sides, so the sum of its interior angles is

$(n-2)\times 180^\circ = (5-2)\times 180^\circ = 540^\circ.$

Because the pentagon is regular, all five interior angles are equal:

$\text{interior angle} = \frac{540^\circ}{5} = 108^\circ.$

So the interior angle of the pentagon at $B$ is $\angle ABC = 108^\circ$.

Step 2: Use the straight line $ABF$

The points $A$, $B$ and $F$ lie on a straight line, so $\angle ABC$ and $\angle CBF$ are angles on a straight line and must add up to $180^\circ$:

$\angle CBF = 180^\circ - \angle ABC = 180^\circ - 108^\circ = 72^\circ.$

Alternative: exterior-angle method

The sum of the exterior angles of any polygon is $360^\circ$. For a regular pentagon, each exterior angle is

$\frac{360^\circ}{5} = 72^\circ.$

Since $BF$ is the extension of $AB$, the angle $CBF$ is exactly the exterior angle of the pentagon at $B$, so $\angle CBF = 72^\circ$ straight away, matching Step 2.

$\boxed{\angle CBF = 72^\circ}$

The correct option is C.