Cyclic Quadrilateral Problem Analysis

The question asks us to find the difference between two specific angles, $\angle DPC$ and $\angle BQC$, formed by producing the sides of a given cyclic quadrilateral ABCD. We are provided with two angles of the quadrilateral: $\angle A = 60^\circ$ and $\angle ABC = 72^\circ$.

Key Properties Used:

• A cyclic quadrilateral is a quadrilateral whose vertices all lie on a single circle.
• A fundamental property of cyclic quadrilaterals is that the sum of opposite angles is $180^\circ$.
• When finding angles in triangles formed by extending sides, we often use the property that angles on a straight line sum to $180^\circ$.

We need to determine the measures of $\angle C$ and $\angle D$ first, and then use the properties of triangles formed by the intersecting lines.

Calculating Unknown Angles of the Quadrilateral

Since ABCD is a cyclic quadrilateral:

• Opposite angles sum to $180^\circ$.
• Therefore, $\angle BCD = \angle C = 180^\circ - \angle A$.
• Substituting the given value of $\angle A$: $$ \angle C = 180^\circ - 60^\circ = 120^\circ $$
• Similarly, $\angle ADC = \angle D = 180^\circ - \angle ABC$.
• Substituting the given value of $\angle ABC$: $$ \angle D = 180^\circ - 72^\circ = 108^\circ $$

So, the angles of the cyclic quadrilateral are $\angle A = 60^\circ$, $\angle ABC = 72^\circ$, $\angle C = 120^\circ$, and $\angle D = 108^\circ$.

Finding the Measure of Angle DPC

Angles $\angle DPC$ is formed at point P, where the extensions of sides AD and BC meet.

Consider the triangle $\triangle PDC$. The sum of angles in a triangle is $180^\circ$. The angles in $\triangle PDC$ are $\angle DPC$, $\angle PDC$, and $\angle PCD$.

• The angle $\angle PDC$ is formed by the line segment PD (extension of AD) and the side DC. Since A, D, P are collinear, $\angle PDC = 180^\circ - \angle ADC$. $$ \angle PDC = 180^\circ - 108^\circ = 72^\circ $$
• The angle $\angle PCD$ is formed by the line segment PC (extension of BC) and the side CD. Since B, C, P are collinear, $\angle PCD = 180^\circ - \angle BCD$. $$ \angle PCD = 180^\circ - 120^\circ = 60^\circ $$
• Now, apply the angle sum property in $\triangle PDC$: $$ \angle DPC + \angle PDC + \angle PCD = 180^\circ $$ $$ \angle DPC + 72^\circ + 60^\circ = 180^\circ $$ $$ \angle DPC + 132^\circ = 180^\circ $$ $$ \angle DPC = 180^\circ - 132^\circ = 48^\circ $$

Finding the Measure of Angle BQC

Angle $\angle BQC$ is formed at point Q, where the extensions of sides DC and AB meet.

Consider the triangle $\triangle QBC$. The sum of angles in this triangle is $180^\circ$. The angles in $\triangle QBC$ are $\angle BQC$, $\angle QBC$, and $\angle BCQ$.

• The angle $\angle QBC$ is formed by the line segment QB (extension of AB) and the side BC. Since A, B, Q are collinear, $\angle QBC = 180^\circ - \angle ABC$. $$ \angle QBC = 180^\circ - 72^\circ = 108^\circ $$
• The angle $\angle BCQ$ is formed by the line segment QC (extension of DC) and the side CB. Since D, C, Q are collinear, $\angle BCQ = 180^\circ - \angle BCD$. $$ \angle BCQ = 180^\circ - 120^\circ = 60^\circ $$
• Now, apply the angle sum property in $\triangle QBC$: $$ \angle BQC + \angle QBC + \angle BCQ = 180^\circ $$ $$ \angle BQC + 108^\circ + 60^\circ = 180^\circ $$ $$ \angle BQC + 168^\circ = 180^\circ $$ $$ \angle BQC = 180^\circ - 168^\circ = 12^\circ $$

Final Calculation: The Difference Between Angles

The question asks for the value of $\angle DPC - \angle BQC$.

• Using the calculated values: $$ \angle DPC - \angle BQC = 48^\circ - 12^\circ $$
• $$ \angle DPC - \angle BQC = 36^\circ $$

Therefore, the difference between the angles $\angle DPC$ and $\angle BQC$ is $36^\circ$. This corresponds to Option 2.