Question

Part I: Using the relationship youestablished in Part I the arcABC = 280 find the mAG. Part Ill: Using vour answer from Part 2 findthe mABC

Answer 1

Given that

`m\angle ABC`

is equal to one half of

`\text{m}\hat{\text{AC}}`

we can set the following equation:

`\text{m}\hat{\text{AC}}=2m\angle ABC.`

Now, substituting the given value for m∠ABC, we get:

`\text{m}\hat{\text{AC}}=2*40^(\circ)=80^(\circ).`

Finally, notice that:

`m\hat{\text{ABC}}=360^(\circ)-m\hat{AC}.^{}`

Therefore:

`\text{m}\hat{\text{ABC}}=280^(\circ).`

Answer:

Part I:

`\text{arcAC}=2*\angle B.`

Part II:

`\text{m}\hat{\text{AC}}=80^(\circ).`

Part III:

`\text{m}\hat{\text{ABC}}=280^(\circ).`