O is the centre of the circle and ABC and EDC are rangers to the circle. Find the side of angle BCD. You must give reason in your answer. (4marks)PLZZZZZZ HELP!!

Every line from the tangent to the centre is 90°. This is said in the circle theorems.
So angle OBC and ODC are 90°.
angle O is twice the size of angle F. This is another circle theorem.
Therefore angle O is 152°.
The quadrilateral of ODBC would equal to 360°. Which means 90° + 90° + 152° = 332°
360° - 332° = 28°.
The answer is 28°
Answer:
∠BCD = 28°
Step-by-step explanation:
arc BD = 2 x inscribed angle ∠BFD = 2 x 76 = 152
arc BFE = 360 - arc BD = 360 - 152 = 208
tangent-tangent angle ∠BCD = (arc BFE - arc BD)/2 = (208 - 152)/2 = 28