ABCD are points on the circumference of a circle with center O. PD is a tangent to the circle at D. Given that $\angle ADB = 28^\circ$ and $\angle CBD = 47^\circ$, calculate the following angles: 1) $\angle BAD$ 2) $\angle CDP$ 3) $\angle CAB$ 4) $\angle BCD$

1. Problem Description

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

1) $\angle BAD$

2) $\angle CDP$

3) $\angle CAB$

4) $\angle BCD$

2. Solution Steps

1) Calculation of $\angle BAD$:

Angles subtended by the same chord at the circumference are equal. Therefore, $\angle CAD = \angle CBD = 47^\circ$.

Also, $\angle BAD = \angle CAD + \angle BAD$.

We are given $\angle ADB = 28^\circ$

Angles subtended by the same chord at the circumference are equal. Therefore, $\angle ACB = \angle ADB = 28^\circ$

Since $\angle BAD$ and $\angle BCD$ are opposite angles in a cyclic quadrilateral $ABCD$, $\angle BAD + \angle BCD = 180^\circ$.

$\angle ABD = \angle ACD$ (Angles subtended by the same chord)

$\angle BAD = \angle CAD + \angle DAB$

$\angle BAD = \angle CAD + \angle BDC = 47^\circ + 28^\circ = 75^\circ$

2) Calculation of $\angle CDP$:

The angle between a tangent and a chord at the point of contact is equal to the angle in the alternate segment.

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

3) Calculation of $\angle CAB$:

$\angle CAB = \angle CDB$ since they subtend the same arc.

Since $\angle CBD = 47^\circ$ and $\angle CDB = \angle CAB$,

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

In triangle $BCD$, we have $\angle CBD = 47^\circ$. We also have that $\angle BDC$ and $\angle BAC$ subtend the same chord, and therefore, $\angle CAB = \angle CDB$.

In the cyclic quadrilateral $ABCD$, $\angle BCD + \angle BAD = 180^\circ$

We know that $\angle BAD = 75^\circ$. Therefore, $\angle BCD = 180^\circ - 75^\circ = 105^\circ$

$\angle BCD = \angle BCA + \angle ACD$

$\angle ACD = \angle ABD$ (Angles subtended by same chord $AD$)

In $\triangle ABD$, we have $\angle BAD = 75^\circ$, $\angle ADB = 28^\circ$, so $\angle ABD = 180^\circ - 75^\circ - 28^\circ = 77^\circ$.

Thus $\angle ACD = 77^\circ$.

Since $\angle BCD = \angle BCA + \angle ACD$, we have $105^\circ = 28^\circ + \angle ABC$

Angles in same segment are equal, so $\angle CAD = \angle CBD = 47^\circ$.

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

In quadrilateral $ABCD$, $\angle BCD + \angle BAD = 180^\circ$.

We have $\angle BAD = 75^\circ$, so $\angle BCD = 105^\circ$.

Since $\angle ABC + \angle ADC = 180^\circ$, $\angle ADC = \angle ADB + \angle BDC$

$\angle CAB = \angle CDB$. Also, $\angle ADB = 28^\circ$ and $\angle CBD = 47^\circ$. Also, $\angle BCD = 105^\circ$.

Consider triangle $BCD$, $\angle CDB = 180 - (47+105) = 180 - 152 = 28^\circ$.

Therefore, $\angle CAB = 28^\circ$.

4) Calculation of $\angle BCD$:

In a cyclic quadrilateral, opposite angles add up to $180^\circ$. Therefore, $\angle BCD + \angle BAD = 180^\circ$.

Since we have calculated $\angle BAD = 75^\circ$, we have $\angle BCD = 180^\circ - 75^\circ = 105^\circ$.

3. Final Answer

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

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

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

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