ABCD are points on the circumference of a circle, with center O. PD is a tangent at D. Angle $ADB = 28^{\circ}$ and angle $CBD = 47^{\circ}$. We need to calculate: (i) Angle $BAD$ (ii) Angle $CDP$ (iii) Angle $CAB$ (iv) Angle $BCD$

1. Problem Description

ABCD are points on the circumference of a circle, with center O. PD is a tangent at D. Angle $ADB = 28^{\circ}$ and angle $CBD = 47^{\circ}$. We need to calculate:

(i) Angle $BAD$

(ii) Angle $CDP$

(iii) Angle $CAB$

(iv) Angle $BCD$

2. Solution Steps

(i) To find angle $BAD$:

Since angles in the same segment are equal, $\angle CAD = \angle CBD = 47^{\circ}$.

Also, we are given that $\angle ADB = 28^{\circ}$.

Then $\angle BAD = \angle CAD + \angle DAB = 47^{\circ} + 28^{\circ} = 75^{\circ}$.

(ii) To find angle $CDP$:

The angle between a tangent and a chord is equal to the angle in the alternate segment. Therefore, $\angle CDP = \angle CAD = 47^{\circ}$.

(iii) To find angle $CAB$:

Since $OC = OB$ (radii of the circle), triangle $BOC$ is isosceles, with $\angle OBC = \angle OCB$.

$\angle BOC = 2 \times \angle BAC$ (angle at the center is twice the angle at the circumference).

$\angle BAC = \angle BAD - \angle CAD = 75^{\circ} - 47^{\circ}= 28^{\circ}$.

Therefore $\angle BOC = 2 \times 28^{\circ} = 56^{\circ}$.

$\angle OBC = \angle OCB = (180^{\circ} - 56^{\circ})/2 = 124^{\circ}/2 = 62^{\circ}$.

$\angle ABC = \angle ABD + \angle DBC = \angle ABD + 47^{\circ}$.

Since $\angle ADB = 28^{\circ}$, then $\angle ACB = \angle ADB = 28^{\circ}$ as they are angles in the same segment.

Another approach:

Since $OA = OB$, $\angle OAB = \angle OBA$.

The angles in quadrilateral ADBC add up to $360^{\circ}$.

We have $\angle BCD = 180^{\circ} - \angle BAD = 180^{\circ} - 75^{\circ} = 105^{\circ}$.

$\angle BCD = \angle BCA + \angle ACD = \angle ADB + \angle ACD = 28^{\circ} + \angle ACD$.

So, $\angle ACD = 105^{\circ} - 28^{\circ} = 77^{\circ}$.

We know that the angle subtended by an arc at the centre is twice the angle subtended at the circumference. So $\angle BOD = 2\angle BAD = 2 \times 75 = 150$.

However we also know $\angle CAD = 47$, hence $\angle COD = 2\angle CAD = 94$.

Now consider the quadrilateral $AOBD$, we see that $\angle ABD = 180 - \angle ACD$. Then $\angle ABC = \angle ABD + 47$ and $BOC = 56$, meaning $BCO = OCB = (180 - 56)/2= 62$

$\angle CAB = 47- (\angle ACB = 28)$

Consider triangle $ABC$, then $\angle BAC + \angle ABC + \angle ACB = 180$, meaning

$\angle CAB + (\angle ABD + \angle DBC) + \angle ADB = 180$, hence

$\angle CAB + (\angle ABD + 47) + 28=180$. Also know that $\angle CAD = 47$ and $\angle DAB = 28$, and $\angle CAB = \angle BAD - \angle DAB - \angleCAD = \angle 75- 47 - angle??$.

Since $\angle BCD = 105^{\circ}$. We have $\angle ACD = \angle BCD - \angle BCA = 105^{\circ} - 28^{\circ} = 77^{\circ}$.

Now we want to find $\angle CAB$. $\angle CAB = \angle CAD - \angle BAD = 75^{\circ} - 47^{\circ}$. Thus $\angle CAB = 28^{\circ}$.

(iv) To find angle $BCD$:

As ABCD is a cyclic quadrilateral, the opposite angles add up to $180^{\circ}$.

Therefore $\angle BCD + \angle BAD = 180^{\circ}$.

$\angle BCD = 180^{\circ} - \angle BAD = 180^{\circ} - 75^{\circ} = 105^{\circ}$.

3. Final Answer

(i) $\angle BAD = 75^{\circ}$

(ii) $\angle CDP = 47^{\circ}$

(iii) $\angle CAB = 28^{\circ}$

(iv) $\angle BCD = 105^{\circ}$