The problem consists of two parts. Part (a) asks to find the measures of angle $Q\hat{A}B$ and angle $B\hat{A}R$ given that the angle between line $PQ$ and the line $AQ$ is $63^{\circ}$. Part (b) asks for the special name of the cyclic quadrilateral $AQBR$.

1. Problem Description

The problem consists of two parts. Part (a) asks to find the measures of angle $Q\hat{A}B$ and angle $B\hat{A}R$ given that the angle between line $PQ$ and the line $AQ$ is $63^{\circ}$. Part (b) asks for the special name of the cyclic quadrilateral $AQBR$.

2. Solution Steps

(a) (i) To find angle $Q\hat{A}B$:

Since the line $PQ$ is a tangent to the circle at point $Q$, the angle between the tangent and the chord $AQ$ is equal to the angle in the alternate segment, which is the angle $Q\hat{B}A$. Thus, $Q\hat{B}A = 63^{\circ}$.

Also, $OA = OB = OQ$ because they are radii of the circle. Since $OA=OQ$, triangle $OAQ$ is an isosceles triangle. Also, angle $OQA = 63^{\circ}$. Then angle $QAO = OQA = 63^{\circ}$.

Then $Q\hat{A}O = 63^{\circ}$.

Angle $A\hat{O}Q = 180^{\circ} - 63^{\circ} - 63^{\circ} = 54^{\circ}$.

The angle $A\hat{O}B$ is a straight line because $AB$ is a diameter of the circle.

Then $A\hat{O}B = 180^{\circ}$. Also, the angle subtended at the center is twice the angle subtended at the circumference. Therefore, $A\hat{O}B = 2 \times A\hat{Q}B$.

$180^{\circ} = 2 \times A\hat{Q}B$, which implies that $A\hat{Q}B = 90^{\circ}$.

Now we know $A\hat{Q}B = A\hat{Q}O + O\hat{Q}B$. $90^{\circ} = 63^{\circ} + O\hat{Q}B$, so $O\hat{Q}B = 90^{\circ} - 63^{\circ} = 27^{\circ}$.

In triangle $OBQ$, since $OB = OQ$, then $O\hat{B}Q = O\hat{Q}B = 27^{\circ}$.

So, $Q\hat{B}A = 63^{\circ}$. $O\hat{B}Q = 27^{\circ}$. So, $O\hat{B}A + O\hat{B}Q = 63^{\circ}$. Therefore, $O\hat{B}A = 63^{\circ} - 27^{\circ} = 36^{\circ}$.

Since $OA = OB$, then $O\hat{A}B = O\hat{B}A = 36^{\circ}$.

Finally, $Q\hat{A}B = Q\hat{A}O + O\hat{A}B = 63^{\circ} + 36^{\circ} = 99^{\circ}$.

(a) (ii) To find angle $B\hat{A}R$:

$ABRS$ is a straight line because $AB$ is the diameter of the circle and $RS$ is tangent to the circle at point $R$.

$B\hat{R}S$ is equal to $B\hat{A}R$.

The measure of $A\hat{B}R$ is $90^{\circ}$ since it is subtended by the diameter.

$ARBS$ is a cyclic quadrilateral. So $A\hat{B}R+A\hat{S}R = 180^{\circ}$. Also, $B\hat{A}R+B\hat{S}R = 180^{\circ}$.

$O\hat{A}B = O\hat{B}A = 36^{\circ}$.

Also, $O\hat{A}R = O\hat{R}A$.

We have $A\hat{O}R = 2 A\hat{B}R$. $R\hat{B}A = R\hat{Q}A$.

$AB$ is the diameter of the circle. $ARBS$ is not necessarily cyclic quadrilateral because $AB$ and $RS$ are parallel to each other.

$B\hat{R}S = B\hat{A}R$.

Since $RS$ is tangent to the circle at $R$, $O\hat{R}S = 90^{\circ}$.

$O\hat{R}A + B\hat{R}O = 90^{\circ}$.

Since $OB = OR$, triangle $OBR$ is an isosceles triangle. Therefore $O\hat{B}R = O\hat{R}B$.

$A\hat{O}B$ is a straight line and thus $180^{\circ}$. $A\hat{O}B = A\hat{O}R + R\hat{O}B$.

We have $2(A\hat{B}R) = A\hat{O}R$ and $2(A\hat{Q}R) = A\hat{O}R$.

$A\hat{B}R = \frac{1}{2}A\hat{O}R$.

Also $A\hat{B}O = 36^{\circ}$.

If $AOB$ is a straight line, then $AOB = 180^{\circ}$.

If $AQBR$ is an isosceles trapezoid, then $AQ || BR$. Then $Q\hat{A}B = A\hat{B}R = 99^{\circ}$.

Then $O\hat{B}R = 36^{\circ}$, which implies $B\hat{A}R = 27^{\circ}$ and $90 -36=54$.

Since $RS$ is parallel to $AB$, then $O\hat{R}S = 90$.

Also $A\hat{O}B = 180^{\circ}$. Then $\angle ROB=180-54 = 126$, and $O\hat{B}R = O\hat{R}B = 27$. Therefore, $B\hat{A}R = 27^{\circ}$.

(b) The cyclic quadrilateral $AQBR$ is an isosceles trapezoid because $AQ || BR$.

3. Final Answer

(a) (i) $Q\hat{A}B = 99^{\circ}$

(ii) $B\hat{A}R = 27^{\circ}$

(b) Isosceles Trapezoid