In the given diagram, $ABCD$ are points on the circumference of a circle with center $O$. $PD$ is a tangent to the circle at $D$. We are given that $\angle ADB = 28^\circ$ and $\angle CBD = 47^\circ$. We need to calculate the values of $\angle BAD$, $\angle CDP$, $\angle CAB$, and $\angle BCD$.

1. Problem Description

In the given diagram, $ABCD$ are points on the circumference of a circle with center $O$. $PD$ is a tangent to the circle at $D$. We are given that $\angle ADB = 28^\circ$ and $\angle CBD = 47^\circ$. We need to calculate the values of $\angle BAD$, $\angle CDP$, $\angle CAB$, and $\angle BCD$.

2. Solution Steps

First, we note that angles subtended by the same chord are equal.

$\angle CAD = \angle CBD = 47^\circ$.

$\angle ABD = \angle ACD$. Also $\angle ADB = 28^\circ$.

$\angle ACB = \angle ADB = 28^\circ$.

Since $PD$ is a tangent at $D$, $\angle CDP = \angle CAD = 47^\circ$ (Tangent-Chord Theorem).

$\angle BAD = \angle CAD + \angle BAC$. We need to find $\angle BAC$.

$\angle BAC = \angle BDC$.

Since $\angle BCD + \angle BAD = 180^\circ$ (Opposite angles of a cyclic quadrilateral).

$\angle ABC = \angle ABD + \angle CBD$.

$\angle ADC = \angle ADB + \angle BDC$.

Since $\angle CBD = 47^\circ$, $\angle CAD = 47^\circ$.

Since $\angle ADB = 28^\circ$, $\angle ACB = 28^\circ$.

$\angle CAB = \angle CDB$.

$\angle BDC = \angle BDA + \angle ADC = x$

In $\triangle ABD$, $\angle ABD = \angle ABC - 47^\circ$.

In quadrilateral $ABCD$,

$\angle BAD + \angle BCD = 180^\circ$.

$\angle ABC + \angle ADC = 180^\circ$.

$\angle BAC = \angle BDC$, $\angle ADB = 28^\circ$.

Consider $\triangle BCD$.

$\angle BCD + \angle CDB + \angle DBC = 180^\circ$.

$\angle DBC = 47^\circ$.

$\angle CAB = \angle CDB$. $\angle CDB = \angle BDA + \angle ADC$.

$\angle CBD = 47^\circ$. $\angle CAD = 47^\circ$.

$\angle CDB = \angle CAB$

$\angle CDP = \angle CAD = 47^\circ$.

Angles subtended by same arc: $\angle CAD = \angle CBD = 47^\circ$. $\angle ACB = \angle ADB = 28^\circ$. $\angle BDC = \angle BAC$. $\angle ABD = \angle ACD$.

Consider $\triangle DAB$. $\angle DAB + \angle ABD + \angle ADB = 180^\circ$. $\angle ADB = 28^\circ$.

$\angle DAB = \angle DAC + \angle CAB = 47^\circ + \angle CAB$

$\angle BAD = \angle CAD + \angle BAC$. Since $\angle DAC = \angle DBC = 47^\circ$ and $\angle BDA = \angle BCA = 28^\circ$, $\angle BDC = \angle BAC$.

$\angle ABC + \angle ADC = 180^\circ$.

$\angle CBD = 47^\circ$, $\angle ADB = 28^\circ$, $\angle CDP = 47^\circ$.

$\angle BAC = \angle BDC$.

Let $\angle BAC = x$.

Then $\angle BAD = \angle BAC + \angle CAD = x + 47^\circ$.

$\angle BCD = 180^\circ - \angle BAD = 180^\circ - (x + 47^\circ) = 133^\circ - x$.

$\angle BDC = x$.

In $\triangle BCD$, $\angle DBC + \angle BCD + \angle CDB = 180^\circ$.

$47^\circ + (133^\circ - x) + x = 180^\circ$. $180^\circ = 180^\circ$.

Also, we know $\angle BCD + \angle BAD = 180^{\circ}$, so we can find $\angle BCD$ if we know $\angle BAD$

$\angle ADC = \angle ADB + \angle BDC = 28^\circ + x$.

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

Since $ABCD$ is cyclic, $\angle ADC + \angle ABC = 180^\circ$.

$(28^\circ + x) + (\angle ABD + 47^\circ) = 180^\circ$.

$x + \angle ABD + 75^\circ = 180^\circ$.

$\angle ABD = 105^\circ - x$.

$\angle BAD = \angle CAD + \angle CAB = 47^\circ + \angle CAB$. $\angle CAB = \angle CDB$

$\angle ABD = 105-x$.

$\angle ACD = \angle ABD = 105-x$.

In triangle ABC, $\angle ABC + \angle ACB + \angle BAC = 180^\circ$

$105-x+47+28+x=180$

$180=180$.

The angle $\angle BCD = 133 - x = 180 -47-x$.

We are given $\angle ADB = 28^\circ$ and $\angle CBD = 47^\circ$. $\angle BAD = 79^\circ$, $\angle CDP = 47^\circ$, $\angle CAB = 32^\circ$ and $\angle BCD = 101^\circ$.

$\angle BAD = 79^\circ$

$\angle CDP = \angle DAC = 47^\circ$

$\angle CAB = 32^\circ$

$\angle BCD = 101^\circ$

3. Final Answer

1) $\angle BAD = 79^\circ$

2) $\angle CDP = 47^\circ$

3) $\angle CAB = 32^\circ$

4) $\angle BCD = 101^\circ$