Geometry Help. Please Help Me!!!!

[kitam965]

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

 

[kitam965]

Im begging you please help me.

 

[Dr.Peterson]

What theorems have you learned that might apply? What have you tried doing? (Did you find arc EH yet? That's the first thing I'd do.)

The main theorem I'd use is on this page: https://mathbitsnotebook.com/Geometry/Circles/CRAngles.html . (Hint: look for mentions of tangents.)

 

[Dr.Peterson]

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]

Given: measure of arc HL =40°, measure of EV =130°, measure of VL =110°.

angle

View attachment 11729

Let's use Prof. Peterson's diagram.
$\displaystyle m(\angle VYE)=\tfrac{1}{2}\{m(arc[EV])-m(arc[HE])\}$

 

[gay]

hey can u pls provide a proof for this

 

[Dr.Peterson]

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.

 

[Steven G]

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]

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: $\displaystyle m(\angle EYH)=\tfrac{1}{2}\left\{m(arc[EV]-arc[EH]\right\}$