ABCD are points on the circumference of a circle with center O. PD is a tangent to the circle at point D. Given that $\angle ADB = 28^\circ$ and $\angle CBD = 47^\circ$, we need to calculate: 1. $\angle BAD$

1. Problem Description

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

1. $\angle BAD$

2. $\angle CDP$

3. $\angle CAB$

4. $\angle BCD$

2. Solution Steps

1. $\angle BAD$:

Since angles subtended by the same arc at the circumference are equal, $\angle CAD = \angle CBD = 47^\circ$.

Therefore, $\angle BAD = \angle CAD + \angle BAD = 47^\circ + 28^\circ = 75^\circ$.

2. $\angle CDP$:

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

3. $\angle CAB$:

Since angles subtended by the same arc at the circumference are equal, $\angle CAB = \angle CDB$.

Also, $\angle CDB = \angle CDA + \angle ADB$, and $\angle CDA = \angle CPA$

We have $\angle CBD = 47^{\circ}$, $\angle ADB=28^{\circ}$. Also $\angle CAD = \angle CBD = 47^{\circ}$. We know that $\angle DAB= \angle DAC + \angle CAB= \angle CBD + \angle ADB = 47+28= 75^{\circ}$

Therefore, $\angle CDB=\angle CAB$. $\angle CDB=\angle CAB=47^{\circ}$

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

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

$\angle CAB = \angle CDB$. Since $ABCD$ is a cyclic quadrilateral, $\angle CDA + \angle CBA = 180^{\circ}$.

$\angle CBA = \angle CBD + \angle DBA$. Also $\angle DCA = \angle DBA$ because angles subtended by arc $AD$ are equal. So $\angle CDA = \angle CDB +\angle BDA$.

Also $\angle CAB = \angle CDB = 47^\circ$

4. $\angle BCD$:

Since $ABCD$ is a cyclic quadrilateral, the sum of opposite angles is $180^\circ$.

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

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

3. Final Answer

1. $\angle BAD = 75^\circ$

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

3. $\angle CAB = 28^{\circ}$

4. $\angle BCD = 105^\circ$

Here is how to calculate the third value correctly.

$\angle CAB = \angle CDB$.

$\angle CDB = \angle CDA + \angle ADB$

Also $\angle DBA = \angle DCA$ (angles in same segment).

In $\triangle BCD$, $\angle CDB = 180 - (\angle CBD + \angle BCD) = 180 - (47 + 105) = 180 - 152 = 28$.

Thus $\angle CAB = 28^\circ$.

Final Answer:

1. $\angle BAD = 75^\circ$

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

3. $\angle CAB = 28^\circ$

4. $\angle BCD = 105^\circ$