In the figure<__ABO__, (AOB, ABO, BOA)
and <__OBC___ (BCO, OBC,BOC) have measures equal to 35°
Measure of an arc of a circle in degrees represents the angle it subtends on the center.
What is the sum of angles of a triangle?
For a triangle ABC, the sum of their internal angles is expressed as
[tex]m\angle A + m\angle B + m\angle C = 180^\circ[/tex]
It is given that for the given figure, we have mCD = 125 degrees.
The arc CD is 125 degrees, thus the angle it subtends on the center is of 125 degrees, or
[tex]m\angle COD = 180^\circ[/tex]
Now, since a straight line has 180 degrees, thus,
[tex]m\angle COD + m\angle DOA = 180^\circ\\m\angle DOA = 180 - 125 = 55^\circ[/tex]
As we have got AB a tangent to the circle O, thus, the radius OB touching the tangent AB is perpendicular to it(it is a theorem that radius of a circle touching tangent is perpendicular to it)
Thus, [tex]m\angle OAB = 90^\circ[/tex]
Then, we have:
[tex]m\angle BOA + m\angle OAB + m\angle ABO = 180^\circ\\m\angle OBA = 180 - 55 - 90 = 35^\circ[/tex]
Thus, [tex]m\angle OBA = m\angle ABO = 35^\circ[/tex]
Let we have
[tex]m\angle OCB = x^\circ\\m\angle OBC = y^\circ[/tex]
Then,
[tex]m\angle OBC + m\angle OCA + m\angle COB = 180^\circ\\x + y + 125 = 180\\ x+y = 55[/tex]
The complete question is that measure of angle ABC is 55 degrees.
Thus, we have:
[tex]\angle ABC = y^\circ + \angle OBA\\55 = y + 35\\y = 20[/tex]
Thus, we have x = 55 - y = 35 or [tex]\angle OCB = x^\circ = 35^\circ[/tex]
Thus,
The complete statement is
In the figure<__ABO__, (AOB, ABO, BOA)
and <__OBC___ (BCO, OBC,BOC) have measures equal to 35°
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