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Sagot :
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
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