Answer:
In the figure ∠ABO and ∠BCO have measures equal to 35°.
Step-by-step explanation:
<u><em>The complete question is</em></u>
For circle O, m CD=125° and m∠ABC = 55°
In the figure<____, (AOB, ABO, BOA) and <_____ (BCO, OBC,BOC) have measures equal to 35°
The picture in the attached figure
step 1
Find the measure of angle COB
we know that
----> by central angle
we have
![arc\ CD=125^o](https://tex.z-dn.net/?f=arc%5C%20CD%3D125%5Eo)
therefore
![m\angle COB=125^o](https://tex.z-dn.net/?f=m%5Cangle%20COB%3D125%5Eo)
step 2
we know that
AB is a tangent to the circle O at point A
so
ABC and ABO are right triangles
In the right triangle ABC
Find the measure of angle BCA
Remember that
---> by complementary angles in a right triangle
we have
![m\angle ABC=55^o](https://tex.z-dn.net/?f=m%5Cangle%20ABC%3D55%5Eo)
substitute
![m\angle BCA+55^o=90^o](https://tex.z-dn.net/?f=m%5Cangle%20BCA%2B55%5Eo%3D90%5Eo)
![m\angle BCA=90^o-55^o=35^o\\](https://tex.z-dn.net/?f=m%5Cangle%20BCA%3D90%5Eo-55%5Eo%3D35%5Eo%5C%5C)
step 3
In the triangle BCO
Find the measure of angle CBO
we know that
---> the sum of the interior angles in any triangle must be equal to 180 degrees
we have
![m\angle COB=125^o](https://tex.z-dn.net/?f=m%5Cangle%20COB%3D125%5Eo)
-----> have measure equal to 35 degrees
substitute
![m\angle CBO+125^o+35^o=180^o](https://tex.z-dn.net/?f=m%5Cangle%20CBO%2B125%5Eo%2B35%5Eo%3D180%5Eo)
![m\angle CBO=180^o-160^o=20^o](https://tex.z-dn.net/?f=m%5Cangle%20CBO%3D180%5Eo-160%5Eo%3D20%5Eo)
step 4
Find the measure of angle ABO
In the right triangle ABO
we know that
----> by angle addition postulate
we have
![m\angle ABC=55^o](https://tex.z-dn.net/?f=m%5Cangle%20ABC%3D55%5Eo)
![m\angle CBO=20^o](https://tex.z-dn.net/?f=m%5Cangle%20CBO%3D20%5Eo)
substitute
![55^o=20^o+m\angle ABO](https://tex.z-dn.net/?f=55%5Eo%3D20%5Eo%2Bm%5Cangle%20ABO)
----> have measure equal to 35 degrees
therefore
In the figure ∠ABO and ∠BCO have measures equal to 35°.