Answer:
θ = 56°
Explanation:
As OA and OB are both radii, triangle AOB is an isosceles triangle:
⇒ m∠ABO = m∠OAB = 48°
Interior angles of a triangle sum to 180°:
⇒ m∠ABO + m∠OAB + m∠AOB = 180°
⇒ 48° + 48° + m∠AOB = 180°
⇒ m∠AOB = 180° - 48° - 48°
⇒ m∠AOB = 84°
Angles on a straight line sum to 180°:
⇒ m∠AOB + m∠DOB = 180°
⇒ 84° + m∠DOB = 180°
⇒ m∠DOB = 180° - 84°
⇒ m∠DOB = 96°
The tangent of a circle is always perpendicular to the radius.
Therefore, tangent BC is perpendicular to radius OB, and tangent DE is perpendicular to radius OD, so:
OBCD is a quadrilateral.
The interior angles of a quadrilateral sum to 360°:
⇒ m∠OBC + m∠BCD + m∠CDO + m∠DOB = 360°
⇒ 90° + θ + 28° + 90° + 96° = 360°
⇒ θ + 304° = 360°
⇒ θ = 360° - 304°
⇒ θ = 56°