aboslutelydestroying

Thanks CPhill for covering up for me

AIB = 134 degrees

Angle IAB + Angle IBA + 134 = 180 degrees

Angle IAB + Angle IBA = 46 degrees

AX and BY are angle bisectors

So IAB = CAB/2

and IBA = CBA/2

(CAB+CBA)/2 = 46

Angle CAB + Angle CBA = 92

That means angle C would be 180 - 92 = 88 degrees

https://web2.0calc.com/questions/help_69801

The answer is (3,-4)

I think this might be a similar question: 

https://web2.0calc.com/questions/angle-bisectors-ax-and-by-of-triangle-abc-meet-at

Are the 2 blanks equal or not?

Thanks so much 

thanks CPhill

Thanks so much CPhill, can't do it without you.

This sum if really a combination of 2 geometric series.

First geometric series: \(1\over{5}\),\(2\over{25}\),\(4\over{125}\), and so on

Second geometric series: \(2\over{5}\), \(4\over{25}\), \(8\over{125}\), and so on

use: \(a_1\over1-r\) formula: The sum of the first geometric series would be: \(1\over{5}\)\(\div\)\(3\over5\)=1/3

2nd would be: 2/5 \(\div\) 3/5 = 2/3

1/3+2/3 = 1

We still didn't add the first term: 1

So 1+1 = 2

I have been working on this problems for a long time, and still did not get the answer