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Given m AED = 35°, mBC = 30, and m F= 65° , what is mAC?

*see picture for more in depth ​

Given m AED = 35°, mBC = 30, and m F= 65° , what is mAC? *see picture for more in-example-1

1 Answer

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Answer:

mAC = 195°

Explanation:

from segment DC, we have that:

mAED + mAEC = 180

35 + mAEC = 180

mAEC = 145°

From the pair of chords crossing, we have that:

mAEC = (mAC + mBD)/2

145 = (mAC + mBD)/2

mAC + mBD = 290 (eq1)

As angle BFD has a tangent point in B and a secant in points D and A, we have:

mBFD = (mAB - mBD)/2

Using mAB = mAC + mBC, we have:

65 = (mAC + mBC - mBD)/2

mAC + mBC - mBD = 130

mAC + 30 - mBD = 130

mAC - mBD = 100 (eq2)

If we sum (eq1) and (eq2), we have:

2mAC = 390

mAC = 195°

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