In the figure AB and CD are two diameters of circle. Interse

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Hi,

Can any1 explain me how angle COB=180 degrees. Why is the sum 180+48?

Hi Karishma

can't we take this way that smaller COB is 132 and then CEB = 180 - (132/2) ? using cyclic quadrilateral property.

All vertices of a cyclic quadrilateral must lie on the circle. Angle COB does not lie on the circle. As given, the figure has no cyclic quadrilateral.

Could anyone please be so kind to explain this question with use of diagrams? I have a very hard time understanding the answer to this question. Thank you very much

we have to connect C to center O and prove that the angle at the center is double the angle CEB.

this is hard.

All angles are measured in degrees.

\(? = x = \angle CEB\)



01. Triangle AOC is isosceles with base AC, hence the 66-degrees angle is justified.

02. Angle BAC is inscribed in the circle, hence the 132-degrees red arc (CEB) is justified.

03. Angle x (our FOCUS) is inscribed in the circle and corresponds to the blue arc CADB, hence:

\(? = {{360 - 132} \over 2} = 114\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( A \right)\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Angle CBE + Angle ODB = 180 degree as sum of opposite angles of a quadrilateral is 180.
So, CBE =180-66=114

karishma,

I have a doubt..so from what I understand is that the angle around O should be 360 degree, so shouldnt angle COB be 132 degree? angle COA and angle AOD adds up to 180 degree. angle BOD is 48 degree, making it all add up to 228 degree. Then should angle COB not be 360-228=132 degree?Where am I going wrong, pls help

2nd doubt- is angle COA =48 degree? if not, why? (i assumed angle BOD and COA to be vertically opposite)

Can uou please tell me how is this inscribed angle. I am nt able to figure it out

Bunuel

Thanksfor the explanantion. It would be great if you could direct me to more questions of inscribed-central angle?

VeritasKarishma & chetan2u

Dear experts,

I thought smaller angle COB is 48 degree. Hence the angle CEB is half of angle COB. Please explain what I am missing here.

Regards,
Arup

VeritasKarishma & chetan2u

Thank you both for the nice explanation. The problem says two diameters intersect at an angle 48 degree. How should I understand which one is 48 among the possible 4 angles?

Regards,
Arup

I really don't understand any of the solutions here. When the diameters intersect at 48 degrees, why do we take COB as 48 and not BOD as 48? Even BOD is an intersection angle right?

Can some pls explain the solution with a diagram? (pls highlight the inscribed and central angle, I am getting confused )

Remember this- an inscribed angle is equal to 1/2 the central angle it intercepts.

Thus, the central angle (excluding COB) is 228 (48*2+132) so divide in two to get CEB

The Angle at <CEB is the Inscribed Angle Subtended by Major Arc CADB


Rule: the Central Angle of a Major Arc is measured by the Reflex Angle around the Center

Rule: the Inscribed Angle Subtended by an Arc = (1/2) * (Central Angle Subtended by that Same Arc)c


Label the Center Point O

we are Given that the Intersection of the 2 Diameters creates:

Angle <COA = 48 deg. = Angle <BOD

the other pairs of Congruent Vertically Opposite Angles that complete the Circle around Point 0:

Angle <COB = 132 deg. = Angle <AOD


Major Arc CABD Subtends:

Inscribed Angle <CEB

and

the Reflex Angle around the Center = <COA + <AOD + <DOB = 48 + 132 + 48 = 228 deg.


Inscribed Angle <CEB = (1/2) * 223 deg. = 114 deg.


-A-

114 deg.

Center angle is twice of inscribed angle ,Yes but position of central angle differs in the case of Minor segment, How come we are not taking COB has central angle in this case?

ScottTargetTestPrep JeffTargetTestPrep Can you help??