The inscribed angle theorem says that

Triangle AOC is isosceles because both AO and CO are radii of the circle and have the same length. This means angles CAO and ACO have the same measure and are congruent.
Angles ACO and COD are congruent because they form an alternating interior pair between the parallel lines AC and OD.
Taking all these facts together, we have

and since angle COB is made up of angles COD and DOB, these angles must be congruent, and so the arcs they subtend (CD and DB, respectively) must also congruent.
195=275-275p
-275 -27
-80=-275p
You do -80 divided by -175 which is 0.45 and your answer is C45%.
Idk but the web always helps me so you can ask that question online
Answer:
sin(A-B) = 24/25
Step-by-step explanation:
The trig identity for the differnce of angles tells you ...
sin(A -B) = sin(A)cos(B) -sin(B)cos(A)
We are given that sin(A) = 4/5 in quadrant II, so cos(A) = -√(1-(4/5)^2) = -3/5.
And we are given that cos(B) = 3/5 in quadrant I, so sin(B) = 4/5.
Then ...
sin(A-B) = (4/5)(3/5) -(4/5)(-3/5) = 12/25 + 12/25 = 24/25
The desired sine is 24/25.
5: AA 47=47 36=36 similar
6: SSS 32/16 =2, 24/12 =2, 9/4.5 = 2 similar
7: SAS 72/53= 1.3, 55=55, 48/12 = 4 not similar
8: SAS 45/15= 3, 54=54, 9/3=3 Similar
9.SAS 20/10 =2, 10/8 = 1.25 not similar
10.SAS 32/14= 2.29, 46=46, 12/7= 1.79 not similar