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 $\angle CBD = 47^{\circ}$, we need to calculate: 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 $\angle CBD = 47^{\circ}$, we need to calculate:

1) $\angle BAD$

2) $\angle CDP$

3) $\angle CAB$

4) $\angle BCD$

2. Solution Steps

1) Calculate $\angle BAD$:

$\angle BAD$ and $\angle BCD$ are angles subtended by the same chord BD in the same segment. Therefore, $\angle BAD = \angle BCD$. Also, $\angle BCD = \angle BCA + \angle ACD$.

$\angle CAD = \angle CBD = 47^{\circ}$ because they are subtended by the same chord CD.

$\angle ACB = \angle ADB = 28^{\circ}$ because they are subtended by the same chord AB.

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

2) Calculate $\angle CDP$:

Since PD is a tangent at D, $\angle CDP = \angle CAD = 47^{\circ}$ (Alternate Segment Theorem).

3) Calculate $\angle CAB$:

$\angle CAB = \angle DAB - \angle DAC = 75^{\circ} - 47^{\circ} = 28^{\circ}$.

Alternatively, $\angle CAB = \angle CDB$. $\angle CDB = \angle CDE$. Also, $\angle CAB=28$.

4) Calculate $\angle BCD$:

$\angle BCD = \angle BAD = 75^{\circ}$ (Angles in the same segment subtended by chord BD).

3. Final Answer

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

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

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

4) $\angle BCD = 75^{\circ}$