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The problem involves a circle with its center at \( O \), and tangent \( CE \) at point \( C \). Let's solve the given sub-questions:

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### **Step-by-step solution:**

#### **a) \(\angle ADO = ?\)**  
- **Given:**
  - \(\angle DAO = 32^\circ\).
  - \( AD \) is a radius of the circle.

- **Key Concept:**
  - In a circle, the angle between a tangent and a radius drawn to the point of tangency is \( 90^\circ \).

- \[
\angle ADO = 90^\circ - \angle DAO = 90^\circ - 32^\circ = 58^\circ
\]

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#### **b) \(\angle ADB = ?\)**  
- **Key Concept:**
  - The angle \(\angle ADB\) is an external angle in triangle \(\triangle ADO\), and it is the sum of the other two internal angles \(\angle DAO\) and \(\angle ADO\).

- \[
\angle ADB = \angle DAO + \angle ADO = 32^\circ + 58^\circ = 90^\circ
\]

---

#### **c) \(\angle DOB = ?\)**  
- **Key Concept:**
  - \(\triangle DOB\) is an isosceles triangle because \( OD = OB \) (both are radii of the circle).
  - The total angle around \( O \) is \( 360^\circ \). Since \(\angle DOB\) is at the center and subtends the arc \( AB \), \(\angle DOB = 2 \times \angle ADB\).

- From part (b):  
  \[
  \angle DOB = 2 \times \angle ADB = 2 \times 90^\circ = 180^\circ
  \]

---

#### **d) \(\angle DBC = ?\)**  
- **Key Concept:**
  - \(\angle DBC\) is an angle formed outside the circle, and \(\angle BOC\) (the central angle at \( O \)) is subtended by arc \( BC \).
  - Using the tangent-segment property and known geometry:
    - \(\angle BDC + \angle DBC = 90^\circ\).

- First, calculate \(\angle BDC\):
  - \(\angle DOB = 180^\circ\), so arc \( AB \) subtends \( 180^\circ \), leaving \( \angle DBC = 90^\circ - \text{(previous symmetry cancellation)}.Sorry, there was a mistake in completing part (d). Let me correct and finalize it carefully.

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### **d) \(\angle DBC = ?\)**  
- **Key Concept:**
  - The angle between a tangent (\(CE\)) and a chord drawn to the point of tangency (\(BC\)) is equal to the angle subtended by the chord on the opposite side of the circle. This is the **alternate segment theorem**.

- The angle subtended by the chord \(BC\) on the opposite side is \(\angle BAD = 32^\circ\). Hence, by the alternate segment theorem:  
  \[
  \angle DBC = \angle BAD = 32^\circ
  \]

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### **Final Answers:**
a) \(\angle ADO = 58^\circ\)  
b) \(\angle ADB = 90^\circ\)  
c) \(\angle DOB = 180^\circ\)  
d) \(\angle DBC = 32^\circ\)  

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### Let me know if you'd like further clarification on any step!  

Here are 5 related questions to expand your understanding:  
1. What is the alternate segment theorem, and how can it be applied in other scenarios involving tangents?  
2. Why is the angle between a radius and tangent always \(90^\circ\)?  
3. How do you calculate angles subtended by arcs at the center and the circumference of a circle?  
4. How would the answers change if \(\angle DAO\) was different?  
5. What is the sum of interior angles in a quadrilateral inscribed in a circle?

### Tip:  
Always check for key circle theorems, such as the tangent-radius angle and alternate segment theorem, as they often simplify complex geometry problems!

In the diagram below, O is the centre of the circle. CE is a tangent to the circle. Solve for the angles: a) ∠ADO, b) ∠ADB, c) ∠DOB, d) ∠DBC.

Explore how to calculate angles in a circle involving tangents, radii, and triangles. Step-by-step solution to determine ∠ADO, ∠ADB, ∠DOB, and ∠DBC using the tangent-radius theorem and alternate segment theorem. Ideal for N5-level math students.

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