#### kitam965

##### New member
Given: measure of arc HL =40°, measure of EV =130°, measure of VL =110°.
Find: m∠EYH.

#### Dr.Peterson

##### Elite Member
Just for fun I drew the figure to scale, and discovered that the problem is wrong -- it's overspecified, with inconsistent data. The three lines don't intersect at Y, but at three points. It should be possible to figure out one of the arcs from the other two, though I haven't worked that out yet.

#### pka

##### Elite Member
Given: measure of arc HL =40°, measure of EV =130°, measure of VL =110°.
Let's use Prof. Peterson's diagram.
$$\displaystyle m(\angle VYE)=\tfrac{1}{2}\{m(arc[EV])-m(arc[HE])\}$$

#### gay

##### New member
hey can u pls provide a proof for this

#### Dr.Peterson

##### Elite Member
Please read our submission guidelines, which apply as much to those who jump into an existing thread as to original posters. Our purpose is not to give complete answers but to help you work things out yourself. So the best thing is for you to show us what you can do with it, and then we can help you along.

Also, please note that I showed that the problem as given is invalid. That means it's not a good one to choose to learn from. And if you ask because you were assigned the same problem, I'll want confirmation that the numbers given are correct.

#### Jomo

##### Elite Member
hey can u pls provide a proof for this
But that would not be helping you. Where exactly are you stuck?
Edit: I now see Dr Peterson says the problem is not valid.

#### pka

##### Elite Member
Given: measure of arc HL =40°, measure of EV =130°, measure of VL =110°.
Find: m∠EYH.View attachment 11727
Using the given: $$\displaystyle m\{arc(EVLH)\}=280^o$$_____ $$\displaystyle \therefore$$$$\displaystyle m\{arc(EH)\}=80^o$$
Now use the tangent-secant angle theorem: $$\displaystyle m(\angle EYH)=\tfrac{1}{2}\left\{m(arc[EV]-arc[EH]\right\}$$

#### Attachments

• 7.6 KB Views: 0