ACB is equal to ADB.I have tried the various theorems and still getting nowhere, i am wondering if the question has enough in it?
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Notice that [imath]\angle ADB~\&~\angle ACB [/imath] are both inscribed angles intercepting the same arc,I'm stuck too! How do you find CAB in the second step?
I get that but still cant see how to find CAB? Apologies for being a dope!Notice that [imath]\angle ADB~\&~\angle ACB [/imath] are both inscribed angles intercepting the same arc,
I can't see how x could equal y unless you make an assumptionI get the same angles as you get. Did you try to draw any additional lines or extend any lines? That might help! How about arc lengths?
Based on you work, how much is x+y equal to?
Does x=y!!!!
Nope. You are absolutely correct. I played with this on Desmos and I came up with three different values for x, depending on where I put D and A.
This suggests its is not a well formed question?Seems like we can move points C and D around the circle while maintaining the 31 degree angles - this changes x.
So you can still retain the 68 and 31 for different x?Nope. You are absolutely correct. I played with this on Desmos and I came up with three different values for x, depending on where I put D and A.
-Dan
Yes. Depending on where I put A after I set point D, I was able to construct x = 11.7 degrees, 12.4 degrees, and 2.5 degrees.So you can still retain the 68 and 31 for different x?
Exactly what I have. x+y=37. But I can't see another relationship between x and y.
I am surly because I can't finish. What are you seeing that we aren't?Can you see that [imath]\Delta AOB[/imath] is isosceles?
Can you see that [imath]m(\angle AOB)=136^o[/imath]?
Can you see that [imath]m(\angle OAB)=22^o[/imath]?
If so, surly you can finish?
[imath][/imath]
[imath][/imath]
[imath][/imath]
[imath][/imath]