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4vir4ik [10]
3 years ago
6

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 y

our answers.

Mathematics
1 answer:
Simora [160]3 years ago
7 0

Answer:

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°

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<img src="https://tex.z-dn.net/?f=e%5E%7B2x%7D%20%3D%20ln%205" id="TexFormula1" title="e^{2x} = ln 5" alt="e^{2x} = ln 5" align=
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Solve : <span>ln(e ^{2x} ) = ln(ln(5))
</span>
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hope this helps!

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