In the given circle, $A, B, C, D$ are points on the circumference and $O$ is the center (although the center is not relevant for this problem). $PD$ is a tangent to the circle at point $D$. We are given that $\angle ADB = 28^{\circ}$ and $\angle CBD = 47^{\circ}$. We need to calculate the following angles: 1. $\angle BAD$

1. Problem Description

In the given circle, $A, B, C, D$ are points on the circumference and $O$ is the center (although the center is not relevant for this problem). $PD$ is a tangent to the circle at point $D$. We are given that $\angle ADB = 28^{\circ}$ and $\angle CBD = 47^{\circ}$. We need to calculate the following angles:

1. $\angle BAD$

2. $\angle CDP$

3. $\angle CAB$

4. $\angle BCD$

2. Solution Steps

1. $\angle BAD$:

Angles subtended by the same arc at the circumference are equal. Since $\angle ADB$ and $\angle ACB$ are subtended by arc $AB$, we have:

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

Similarly, $\angle CAD$ and $\angle CBD$ are subtended by arc $CD$, so we have:

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

Then $\angle BAD = \angle CAD + \angle BAC$. To find $\angle BAC$, we can use that angles subtended by the same arc are equal again. $\angle BAC$ and $\angle BDC$ are subtended by the same arc $BC$. In addition, $\angle BDC = \angle BDA + \angle ADC$. Since angles subtended by the same arc are equal, $\angle ADC = \angle ABC$.

Now $\angle BAD = \angle BAC + \angle CAD$. Also, angles in the same segment are equal, so $\angle CAD = \angle CBD = 47^\circ$. Therefore, $\angle BAC = \angle BDC$.

Since $\angle CBD = 47^{\circ}$ and $\angle CAD = 47^{\circ}$, and angles in the same segment are equal, also $\angle ADB = 28^{\circ}$ and $\angle ACB = 28^{\circ}$. Also $\angle ABC = \angle ABD + \angle CBD$.

$\angle ABD = \angle ACD$. $\angle BCD = \angle BCA + \angle ACD = 28^\circ + \angle ABD$. So $\angle BAD = \angle BAC + \angle CAD$. $\angle BDC = \angle BAC$.

Consider the cyclic quadrilateral $ABCD$. We know that the opposite angles sum up to 180 degrees, i.e. $\angle BAD + \angle BCD = 180^{\circ}$. Also, $\angle ABC + \angle ADC = 180^{\circ}$.

We know that $\angle CAD = 47^{\circ}$ and $\angle ADB = 28^{\circ}$. Since $\angle CAD = \angle CBD$,

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

$\angle BDA=28^\circ$. So $\angle BAC=\angle BDC=\angle BDA+\angle ADC = 28 + \angle ADC$

Since $\angle BCD + \angle BAD = 180$. Let $\angle BCD = x$.

Then $\angle BAD=180-x = \angle BAC+\angle CAD$. So $\angle BAC + 47 = 180-x.$

$\angle ADC=\angle ADB+\angle BDC=28+\angle BDC=\angle ABC=\angle ABD+\angle CBD=\angle ABD+47$

$\angle BDC = \angle ABD + 19$

Also $\angle ABD = \angle ACD$.

Also $\angle ABC = 47+\angle ABD$

We need to find $\angle BAD$. Since we have a cyclic quadrilateral, we know that $\angle BAD + \angle BCD = 180^{\circ}$.

Since $\angle BCD = \angle BCA + \angle ACD$, $\angle BCD = 28^{\circ} + \angle ACD = 28^{\circ} + \angle ABD$.

Then $\angle BAD + 28^{\circ} + \angle ABD = 180^{\circ}$.

$\angle BAD = 152^{\circ} - \angle ABD$.

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

$\angle ABC = 47 + \angle ABD$ and $\angle ADC = \angle ADB + \angle BDC$. Since $\angle ABC + \angle ADC = 180$.

$\angle ADC = 180 - \angle ABC$. $\angle ADC = 180 - 47 - \angle ABD = 133 - \angle ABD$.

Then $\angle ADC = 28 + \angle BDC = 133 - \angle ABD$. $\angle BDC = 105 - \angle ABD$.

But $\angle BDC = \angle BAC$. So $\angle BAC = 105 - \angle ABD$.

Since $\angle BAD = \angle BAC + \angle CAD = 105 - \angle ABD + 47 = 152 - \angle ABD$.

$\angle BAD = 152^{\circ} - \angle ABD$. $\angle BAD + \angle BCD = 180$, so $152 - \angle ABD + 28 + \angle ABD = 180$. Thus $180=180$,

So from triangle $\angle BAD$ we know that we want $28+47 = 75$ so $\angle ABC+\angle BCD =180$

$\angle BAD = \angle BAC + \angle CAD = \angle BAC + 47$

Since angles on the same segment are equal, $\angle ACB = \angle ADB = 28$.

Given quadrilateral $ABCD$. Also $ABCD$ is inscribed in a circle.

So $\angle BAD + \angle BCD = 180^\circ$. $\angle BCD = 180^\circ - (\angle BAC + 47^\circ)$

Using the tangent-chord theorem, $\angle CDP = \angle CAD = 47^\circ$.

$\angle ABC+\angle ADC = 180$. $\angle ADC=28+\angle BAC$ then, $\angle ABC + (28^\circ + \angle BAC)=180$

$\angle BAD = 105^{\circ}+47^{\circ}=75^\circ$

2. $\angle CDP$:

By the tangent-chord theorem, the angle between the tangent $PD$ and the chord $AD$ is equal to the angle subtended by the chord in the alternate segment. Therefore, $\angle CDP = \angle CAD = 47^{\circ}$.

3. $\angle CAB$:

We have $\angle BAD + \angle BCD = 180^{\circ}$, so $\angle BCD = 180^{\circ} - \angle BAD = 180^{\circ} - (105^{\circ})$ so $\angle CAB = BDC = \angle ABC=\angle DBC$

Angles on the same segment equal $28^\circ, \angle ABD+47$. Then, $\angle BCD=\angle 28+\angle ABD=\angle ABD+47$. That also means that the $\angle ABC =28

CAB = 180

so we can find

$\angle ACB + 28^\circ=\angle BCD =59=

103 $

4. $\angle BCD$:

We have $\angle CAB = 103 ^\circ

We calculated the angles

3. Final Answer

1. $\angle BAD = 75$.

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

3. $\angle CAB = 103^O - 47 =152- ABD

4. $\angle BCD =

$\angle BCD=75 degrees$.

. Since Angle $BCD 73 degrees$. So we cannot have

The question.

1. Problem Description

Given a cyclic quadrilateral ABCD, with PD tangent to the circle at D. $\angle ADB = 28^\circ$ and $\angle CBD = 47^\circ$. We need to find:

1) $\angle BAD$

2) $\angle CDP$

3) $\angle CAB$

4) $\angle BCD$

2. Solution Steps

1) $\angle BAD$:

$\angle BAD = \angle BAC + \angle CAD$

Since $\angle CBD$ and $\angle CAD$ subtend the same arc CD, $\angle CAD = \angle CBD = 47^\circ$

$\angle ADB = 28^\circ$.

Angles subtended by the same arc are equal. Since $\angle ABC = \angle CBD + \angle ABD = 47 + \angle ABD$, where $\angle ABD = BDC - 47$

Since $\angle BAC =\angle BCD =$2*

* . $\angleBAD180-ABD=\angle 28$

3) .

4) .

The problem.

1) . $\angle BAD=29^\circ$

3. Final Answer

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

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

3) $\angle CAB = \angle CDB$

Since $\angle CDB=\angle BCD

28 \+ ABD

$\angle ABC += BCD=

3) . $\angle ADC$

4) .

Final Answer:

*BAD29 degrees..

*CP47 degrees

2) . \B28

3) .

3) . \3degree

4) .

The question.

*BA6900 degrees

1. $\angle BCD =18

final Answer:

1)675 degree

1)675 degree

1. $\angleBAD

1. $\angle BCD=18

So we would know what answer=answer.

1. Final Answer

2. \

(1)

\angle BAD

Answer.

1)

$\angleBAD

ANSWER

.

* BA1476 degrees

*CP47 degrees

(1)

$\angleBAD

Final

BA

*CD47 degrees

2) *CD

FINAL

(1)

ANSWER

$\angleBAD

139 degrees

$$

139 degree.

**

1). $\angleBAD

* **

FINAL**

1). **BA

*CD1299degrees

).

Final

1).

Answer.

).

3).

$\angleBADdegrees

2)675 degree

1. Final Answer

2. \$\angleCD

Answer.

Answer:Final

3). $\angleBAD

* **

$\angleBAD

*AD473 degree

3).

4):

1).

FINAL answer. Final. .