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O is the centre of the circle and ABC and EDC are tangents to the circle

User Duske
by
4.2k points

2 Answers

6 votes

Answer:

28

Explanation:

The angle mBFD inscribes the arc mBD, so we have that:

mBFD = mBD/2

76 = mBD/2

mBD = 152°

The angle mBOD is a central angle related to the arc mBD, so we have that:

mBOD = mBD = 152°

In the quadrilateral BODC, the sum of internal angles needs to be equal to 360° (property of all convex quadrilaterals). The angles mCBO and mCDO are right angles, because EDC and ABC are tangents to the circle.

So, we have that:

mBOD + mCDO + mBCD + mCBO = 360

152 + 90 + mBCD + 90 = 360

mBCD = 360 - 90 - 90 - 152

mBCD = 28°

User Ram Singh
by
4.0k points
5 votes

Answer:

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