Question

Can someone please help with these problems I’m having trouble pls

Answer 1

When the chords intersect inside the circle the sum of the vertical angles and the sum of the referenced arcs are the same.  Since the vertical angles are congruent, each is half the sum of the referenced arcs.

Similarly when the chords intersect outside the circle, the angle is half the difference of the arc measures.

It's the same with a tangent; in that case there's just not a part of the arc that's not counted on the side with the tangent.

OK, that's how.  Let's try some.

1.  x = (88 + 86)/2 = 87

2. x = (150 - 60)/2 = 45

3. The angle is exactly half the subtended arc,

x = 110/2 = 55

4. The referenced arc is 360 - 220 = 140 degrees and x is half, 70.

5.  Outside, half the difference, x = (90-20)/2 = 35

6.  Angle outside, inner arc 6 degrees, vertical angle is 90 degrees.

90 = (x - 6)/2

180 = x - 6

x = 186

7.  There's a special rule for this case but we can treat it as outside, little arc x, big arc 360-x,

60 = (360 - x  - x)/2

120 = 360 - 2x

2x = 240

x = 120

8. Inner arc x, outer arc is 360-140-x = 220-x.  Half the difference is 38,

38 = (220 - x - x)/2

76 = 220 - 2x

2x = 220 - 76

x = (220 - 76)/2 = 72

9.  x is formed by a tangent and a radius, so a right angle, x=90

y is subtended by a 35 degree angle so y=70

10.  Remaining triangle angle is 180 - 62 - 28 = 90.  Arc that subtends it is double, 180.  That means the long side is a diameter.

11.  Remaining triangle angle is 180 - 50 - 70 = 60. x=120, y=100, z=140

12.  Remaining arc 360 - 170 - 90 = 100.  Angles are half.

x=45, y=50, z=85