The intersecting secant-tangent theorem says that
where T is the point at which line segment PO touches the circle. Similarly,
so we have
OP is of course as long as itself, and AO and BO are radii of the same circle so they have the same lengths.
This means triangles APO and BPO are congruent, which means angles APO and BPO are congruent, so angle APB is bisected by PO.
You would have to find Circumference
Answer: A, C, D and E
Step-by-step explanation:
The opposite angles of a parallelogram are equal.
Therefore
m∠A = m∠C, so that
5y - 3 = 3y + 27
Subtract 3y from each side.
5y - 3y - 3 = 3y - 3y + 27
2y - 3 = 27
Add 3 to each side.
2y - 3 + 3 = 27 + 3
2y = 30
y = 30/2 = 15
Therefore
m∠A = 5*15 - 3 = 72°
m∠C = 72°
Let x = m∠B
Then x = , m∠B = m∠C
Because the sum of the angles in the parallelogram is 360°, therefore
x + x + 72 + 72 = 360
2x = 360-144 = 216
x = 216/2 = 108
Answer:
m∠A = 72°
m∠B = 108°
B is one of them, i'm not sure how any of the other answers are equivalent to 20% of 45 which is 8.