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

Question

Circle_Angle.JPG In the figure AB and CD are two diameters of circle. Intersecting at angle 48 degree. E is any point on Arc CB. find angle CEB A. 114 B. 100 C. 80 D. 96 E. 40

KarishmaB · Answer

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.

saiesta · Answer

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

thangvietnam · Answer

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

this is hard.

jaysonbeatty12 · Answer

Quick elimination tip here

An triangle formed by the diameter of a circle and any other point on that circle is a right triangle. The diameter is the hypotenuse and the right angle is formed by the two legs extending to the point. So, in this case, we can create a right triangle by drawing a line from, say, E to D. This shows that angle CED is greater than 90, so eliminate C, D and E.

fskilnik · Answer

All angles are measured in degrees.

? = x = \angle CEB

https://preview.ibb.co/ejhtQA/12-Nov18-4w.gif

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.

HAPPYatHARVARD · Answer

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

yashna36 · Answer

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)

GMATGuruNY · Answer

An INSCRIBED ANGLE is formed by two chords.
A CENTRAL ANGLE is formed by two radii.
When an inscribed angle and a central angle intercept the same two points on a circle, the inscribed angle is 1/2 the central angle.
 https://i.postimg.cc/qgbxWKdW/Angle-CEB-1.png 
Here, inscribed angle CFB and central angle COB both intercept the circle at points C and B.
Since central angle COB = 132º, inscribed angle CFB = (1/2)(132) = 66º.

Rule:
When a quadrilateral is inscribed in a circle, opposite angles sum to 180º.
 https://i.postimg.cc/23VZbXPX/Angle-CEB-2.png 
Since opposite angles inside inscribed quadrilateral CEBF must sum to 180º, angle CEB = 180-66 = 114.

A

AlN · Answer

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

AlN · Answer

Bunuel

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

ArupRS · Answer

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

chetan2u · Answer

No COB is not 48..
Even if you take COB as 48, CEB is not half of COB..
Any point say r in major arc CB that is CADB will make an angle CrB which will be half of COB.
This becomes half of 48 so 24..
Now CEB+CrB will be 180 as sum of opposite angles of a cyclic quadrilateral is 180 and CEBr is cyclic quadrilateral.
Thus CEB = 180-24=156..
But no choice is given as 156.

So you have to take COB as 180-48=132
And CEB = 180-132/2=180-66=114

KarishmaB · Answer

Angle COB is the larger angle so it will not be 48 degrees.
Also, inscribed angle is half of the central angle that subtends the same arc.

COB and CEB subtend different arcs.
Angle COB subtends arc CEB.
Angle CEB subtends arc CADB. 
So angle COB will be twice of angle CAB or CDB.

ArupRS · Answer

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

mup05ro · Answer

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

)

dcummins · Answer

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

Fdambro294 · Answer

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.

AjitPatil · Answer

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?

NakulDiwakar10 · Answer

ScottTargetTestPrep   JeffTargetTestPrep  Can you help??

bumpbot · Answer

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

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