ABCD are points on the circumference of a circle, center O. PD is a tangent at D. Given that angle ADB = $28^{\circ}$ and angle CBD = $47^{\circ}$, we need to calculate: i) angle BAD ii) angle CDP iii) angle CAB iv) angle BCD

1. Problem Description

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

i) angle BAD

ii) angle CDP

iii) angle CAB

iv) angle BCD

2. Solution Steps

i) To find angle BAD:

Angles in the same segment are equal.

Angle BAD = angle BCD.

Angle BCD = angle BCA + angle ACD.

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

Angle CBD = $47^{\circ}$, and since angle CAD and angle CBD are angles in the same segment, then angle CAD = $47^{\circ}$.

Angle BCD = angle BCA + angle ACD. Since angle ACD = angle ABD, then angle BCD = $28^{\circ} + 47^{\circ}$.

Angle BCD = angle BCA + angle ACD = angle ADB + angle ABD = 28° +47° = 75°

Therefore, Angle BAD = $75^{\circ}$.

ii) To find angle CDP:

Since PD is a tangent to the circle at D, angle CDP = angle CAD (alternate segment theorem).

Since angle CAD = $47^{\circ}$, then angle CDP = $47^{\circ}$.

iii) To find angle CAB:

Angle CAB = angle BAD - angle CAD

Angle CAB = $75^{\circ} - 47^{\circ}$

Angle CAB = $28^{\circ}$.

iv) To find angle BCD:

As already calculated in step (i), angle BCD = $75^{\circ}$.

3. Final Answer

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

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

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

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