very difficult geometry

[logistic_guy]

here is the question

Find $\displaystyle m\overset{\large\frown}{QR}$.




my attemb
i can find $\displaystyle m\measuredangle T$
do this help to find $\displaystyle m\overset{\large\frown}{QR}$?

 

[Dr.Peterson]

Find $\displaystyle m\overset{\large\frown}{QR}$.

View attachment 38960
my attempt
i can find $\displaystyle m\measuredangle T$
do this help to find $\displaystyle m\overset{\large\frown}{QR}$?

No, it is irrelevant. The problem can be solved without S being there at all.

Never call a problem "very difficult"; people will probably end up laughing at you.

Just draw in another line -- one that is very relevant.

 

[pka]

here is the question
Find $\displaystyle m\overset{\large\frown}{QR}$.
View attachment 38960

Did it occur to you that $m(\widehat{TQR})=180^o$?
Hence & therefore $m(\widehat{TQ})+m(\widehat{QR})=m(\widehat{TQR})$!

 

[logistic_guy]

thank Dr.

No, it is irrelevant. The problem can be solved without S being there at all.

Never call a problem "very difficult"; people will probably end up laughing at you.

if i'm advance engineering and i can't solve this question
what's the flaw to call it very difficult?

people will probably end up laughing at you.

do you laugh at me?

Just draw in another line -- one that is very relevant.

i see the idea to draw a line from $\displaystyle T$ to $\displaystyle Q$
it give me triangle but i've no idea what kind of triangle is this and how to deal with it

if you mean i put a point outside the circle and do tricks by connecting the point with the circle like teacher Aion do last time
i don't see any idea there

Did it occur to you that $m(\widehat{TQR})=180^o$?
Hence & therefore $m(\widehat{TQ})+m(\widehat{QR})=m(\widehat{TQR})$!

thank pka

the radius isn't horizontal to say the arc is $\displaystyle 180^{\circ}$ and if i say what you say i've to proof it
i don't know how to do that

 

[Dr.Peterson]

if i'm advance engineering and i can't solve this question
what's the flaw to call it very difficult?

It simply isn't at all difficult, if you know basic geometry.

do you laugh at me?

I avoid calling a problem very difficult myself, because I may well have missed something obvious. This is just general advice.

But I don't laugh at you; I'm sad for you.

i see the idea to draw a line from T to Q
it give me triangle but i've no idea what kind of triangle is this and how to deal with it

if you mean i put a point outside the circle and do tricks by connecting the point with the circle like teacher Aion do last time
i don't see any idea there

My suggestion was to use the same basic idea you have already used, since you clearly know at least that much. Yes, TQ is a useful line to draw. if you know about angles in a circle (as you clearly do in part), then you should see an important fact about it.

the radius isn't horizontal to say the arc is $\displaystyle 180^{\circ}$ and if i say what you say i've to proof it
i don't know how to do that

Do you recognize that TR is a diameter? That is the important fact. (pka's suggestion is simpler than mine.)

 

[logistic_guy]

Do you recognize that TR is a diameter? That is the important fact. (pka's suggestion is simpler than mine.)

i think i see the answer now

$\displaystyle m\overset{\large\frown}{QR} = 180^{\circ} - m\overset{\large\frown}{TQ} = 180^{\circ} - 2m\measuredangle R = 180^{\circ} - 2(50^{\circ}) = 180^{\circ} - 100^{\circ} = 80^{\circ}$

is my analize correct?

 

[Dr.Peterson]

Yes.

Or you could draw in TQ as I suggested, and observe that triangle TQR, being inscribed in a semicircle, is a right triangle, so that angle QTR is 40 degrees, so arc QR is 80 degrees.

 

[logistic_guy]

Yes.

Or you could draw in TQ as I suggested, and observe that triangle TQR, being inscribed in a semicircle, is a right triangle, so that angle QTR is 40 degrees, so arc QR is 80 degrees.

thank Dr. for this idea. i see it before but can't see it's a right triangle

you've great insight

i appreciate your help

 

[Dr.Peterson]

thank Dr. for this idea. i see it before but can't see it's a right triangle

Angle inscribed in a semicircle - Math Open Reference

 

[logistic_guy]

Angle inscribed in a semicircle - Math Open Reference

thank Dr.i see the idea now