Geometry Problem: Finding Angle RQP

We are given a circle with centre O and diameter PQ. A point R is on the circle such that $\angle RPO = 42^\circ$. We need to find $\angle RQP$.

Analyzing Triangle RPO

Consider the triangle $\triangle RPO$. Since OR and OP are both radii of the same circle, they are equal in length ($OR = OP$).

• This means $\triangle RPO$ is an isosceles triangle.
• In an isosceles triangle, the angles opposite the equal sides are equal. Therefore, $\angle ORP = \angle RPO$.
• Given $\angle RPO = 42^\circ$, we have $\angle ORP = 42^\circ$.

Calculating Angle ROP

The sum of angles in any triangle is $180^\circ$. Applying this to $\triangle RPO$:

$\angle ROP + \angle RPO + \angle ORP = 180^\circ$

$\angle ROP + 42^\circ + 42^\circ = 180^\circ$

$\angle ROP + 84^\circ = 180^\circ$

$\angle ROP = 180^\circ - 84^\circ = 96^\circ$

Using Angle in Semicircle Property

PQ is the diameter of the circle. The angle subtended by a diameter at any point on the circumference is a right angle ($90^\circ$).

• Therefore, $\angle PRQ = 90^\circ$.
• Now consider the triangle $\triangle PRQ$. The sum of its angles is $180^\circ$.
• $\angle RPQ + \angle PRQ + \angle RQP = 180^\circ$
• We know $\angle RPQ$ is the same as $\angle RPO$, which is $42^\circ$. We also know $\angle PRQ = 90^\circ$.
• $42^\circ + 90^\circ + \angle RQP = 180^\circ$
• $132^\circ + \angle RQP = 180^\circ$
• $\angle RQP = 180^\circ - 132^\circ$
• $\angle RQP = 48^\circ$

Alternatively, we can use the property that $\angle ROQ$ is the angle at the centre subtended by the arc RQ. $\angle ROP$ and $\angle ROQ$ are supplementary angles as PQ is a diameter.

$\angle ROQ = 180^\circ - \angle ROP = 180^\circ - 96^\circ = 84^\circ$.

In $\triangle ROQ$, OR = OQ (radii), so it is an isosceles triangle.

$\angle ORQ = \angle OQR = \frac{180^\circ - \angle ROQ}{2} = \frac{180^\circ - 84^\circ}{2} = \frac{96^\circ}{2} = 48^\circ$.

Thus, $\angle RQP = 48^\circ$.