1. Angle RPB is 72 degrees
2. Arc POQ is 220 degrees.
3. Angle ABC is 135.5 degrees.
Problem 1: Since AR = AB, triangle ARB is isosceles. Also, arc AB is 130 degrees, so angle ARB is 130/2 = 65 degrees, which means angle RAB is also 65 degrees. Then arc RB is 130 degrees. Since arc AQ is 28 degrees, angle RPB is (130 + 28)/2 = 79 degrees.
Problem 2: Since arc POQ is 80 degrees, angle POQ is 160 degrees. Then major arc PQ is 360 - 160 = 200 degrees, so arc POQ on the smaller circle is (360 - 200) + 80 = 240 degrees.
Problem 3: Angle ABX = Angle AOX = 35 degrees, so Angle ABC = 180 - 35 = 145 degrees.
What have you actually tried?
And why is this so "urgent"? Are you just trying to get us to do your homework for you?
+129899 First one
Connect RB
Then.....since AR = AB..... then in triangle ARB, angle ARB = angle ABR
And if arc AB = 130 then angle ARB = 130/2 = 65° = angle ABR
So angle RAB = angle PAB = 180 -130 = 50°
And since arc AQ = 28, then angle QBA = angle PBA = 14°
So by the exterior angle theorem angle RPB = angle PAB +angle PBA = 50 + 14 = 64°
+129899 Second one
Since in the larger circle, minor arc PQ = 80 degrees
Then central angle POQ also = 80°
But if angle POQ = 80°, then, in the smaller circle..... minor arc PQ = 160°
Then arc POQ = 360 -160 = 200°
