O is the centre of the circle and ABC and EDC are tangents to the circle. Find the size of angle BCD. You must give reasons in your answers.

mBCD = 28°

Step-by-step 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°

step-by-step explanation:

in the absence of parentheses, i will have to make some assumptions:   i interpret your (3x+5/y^6)(4y^3/5)​ to mean:

3x + 5     4y^(3/5)       (3x + 5)(4)

* =

    y^6             1               y^(6-3/5)

if this is not what you expected, repost your question using parentheses to indicate whether you meant

4y^3

    or   4y^(3/5)

    5