(a): $42$

(b): $55$

(c): $85$

(d): $108$

(e): $53$

Detailed Solution

(a)

$\angle$s in the same segment subtended by the same arc are equal.
$\angle$ $CAD$ and $\angle$ $CBD$ subtend the same arc $CD$.
Therefore, $\angle$ $CBD$ = $\angle$ $CAD = 42^{\circ}$.

(b) 

The $\angle$ between a tangent and a chord through the point of contact is equal to the $\angle$ in the alternate segment.
$PQ$ is a tangent at $A$, and $AB$ is a chord.
Therefore, $\angle$ $BAQ$ = $\angle$ $ACB$.
Given that $\angle$ $BAQ = 55^{\circ}$, so $\angle$ $ACB = 55^{\circ}$.

(c) 

$ABX$ is a straight line, so $\angle$ $ABC$ and $\angle$ $CBX$ are linear pairs and their sum is $180^{\circ}$.
$\angle$ $ABC +$ $\angle$ $CBX = 180^{\circ}$
$\angle$ $ABC + 85^{\circ} = 180^{\circ}$
$\angle$ $ABC = 180^{\circ} – 85^{\circ} = 95^{\circ}$.
$ABCD$ is a cyclic quadrilateral, so the sum of opposite $\angle$s is $180^{\circ}$.
$\angle$ $ABC +$ $\angle$ $ADC = 180^{\circ}$
$95^{\circ} +$ $\angle$ $ADC = 180^{\circ}$
$\angle$ $ADC = 180^{\circ} – 95^{\circ} = 85^{\circ}$.

(d) 

In tri$\angle$ $ADC$, the sum of $\angle$s is $180^{\circ}$.
$\angle$ $DAC +$ $\angle$ $ADC +$ $\angle$ $DCA = 180^{\circ}$
$42^{\circ} + 85^{\circ} +$ $\angle$ $DCA = 180^{\circ}$
$127^{\circ} +$ $\angle$ $DCA = 180^{\circ}$
$\angle$ $DCA = 180^{\circ} – 127^{\circ} = 53^{\circ}$.
$\angle$ $BCD =$ $\angle$ $BCA +$ $\angle$ $ACD$
$\angle$ $BCD = 55^{\circ} + 53^{\circ} = 108^{\circ}$.

(e) 

The $\angle$ between a tangent and a chord through the point of contact is equal to the $\angle$ in the alternate segment.
$PQ$ is a tangent at $A$, and $AD$ is a chord.
Therefore, $\angle$ $PAD$ = $\angle$ $ACD$.
$\angle$ $ACD$ (which is the same as $\angle$ $DCA$) is $53^{\circ}$.
So, $\angle$ $PAD = 53^{\circ}$.