Question

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!!

Answer 1

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 2

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