Circle

[goyalsau]

[beat_gmat_09]

BC = BD since tangents are drawn from same point B, they are equal. Triangle BDC is Isosceles triangle with base angle B.
Angle BDC is theta = angle BCD
Angle ACB = 90 .. diameter property.
Angle BAC + Angle ACB + Angle CBA = 180
24 + 90 + Angle CBA = 180
CBA= 180-114 = 66
In triangle BDC,
theta + theta + Angle CBD = 180
2(theta) + 66 = 180
2(theta) = 180-66
2(theta) = 114
Theta = 57
Whats the OA and OE ?

[goyalsau]

OA is 38

[diebeatsthegmat]

OA is 38

how come my answer is E? do you have its explanation?

[goyalsau]

angle ADE = Angle DCE

Because of Alternate Segment Theorem , - if you have any doubts look at the link https://www.youtube.com/watch?v=nwOSsjLDACQ

Lets Assume Angle AED = e

and angle ADE = x,
which means ADE = DCE = EDC = x

in Triangle DEC

x + x + 180 - e = 180

e = 2x


in Triangle AED

24 + e + x = 180

24 + 3x = 180

x = 52

Angle ACB is equal to 90

90 = 52 + theta

theta = 38

[beat_gmat_09]

angle ADE = Angle DCE

Because of Alternate Segment Theorem , - if you have any doubts look at the link https://www.youtube.com/watch?v=nwOSsjLDACQ

Lets Assume Angle AED = e

and angle ADE = x,
which means ADE = DCE = EDC = x

in Triangle DEC

x + x + 180 - e = 180

e = 2x


in Triangle AED

24 + e + x = 180

24 + 3x = 180

x = 52

Angle ACB is equal to 90

90 = 52 + theta

theta = 38

This problem seems absurd, value should be same regardless of approaching problem. I will accept this solution if anyone prove's my approach faulty.

[goyalsau]

BC = BD since tangents are drawn from same point B, they are equal. Triangle BDC is Isosceles triangle with base angle B.
Angle BDC is theta = angle BCD
Angle ACB = 90 .. diameter property.
Angle BAC + Angle ACB + Angle CBA = 180
24 + 90 + Angle CBA = 180
CBA= 180-114 = 66
In triangle BDC,
theta + theta + Angle CBD = 180
2(theta) + 66 = 180
2(theta) = 180-66
2(theta) = 114
Theta = 57
Whats the OA and OE ?

It will be good if you re read the question once again,
Not any where in the question stem it is given that BC is the Tangent to the circle,

[beat_gmat_09]

Not any where in the question stem it is given that BC is the Tangent to the circle,

It has to be a tangent. Angle ACB is 90 degrees, segment B meets the extreme ends of both the circle at C.
This is where this question is flawed.

[goyalsau]

Not any where in the question stem it is given that BC is the Tangent to the circle,

It has to be a tangent. Angle ACB is 90 degrees, segment B meets the extreme ends of both the circle at C.
This is where this question is flawed.

It will be good if you read this.

https://www.beatthegmat.com/three-grades ... tml#317536

[Rahul@gurome]

Not any where in the question stem it is given that BC is the Tangent to the circle,

It has to be a tangent. Angle ACB is 90 degrees, segment B meets the extreme ends of both the circle at C.
This is where this question is flawed.

BC is neither a tangent to the large circle nor to the smaller.
BC would've been the tangent at point C only if joining the radius of the circle and C resulted in a 90° the line and BC. But as AC is perpendicular BC at point C and AC is not a part of radius of any of the circles, BC is not tangent to any of them.

The solution is θ = 38° and method is as goyalsau demonstrated.

[beat_gmat_09]

Not any where in the question stem it is given that BC is the Tangent to the circle,

It has to be a tangent. Angle ACB is 90 degrees, segment B meets the extreme ends of both the circle at C.
This is where this question is flawed.

BC is neither a tangent to the large circle nor to the smaller.
BC would've been the tangent at point C only if joining the radius of the circle and C resulted in a 90° the line and BC. But as AC is perpendicular BC at point C and AC is not a part of radius of any of the circles, BC is not tangent to any of them.

The solution is θ = 38° and method is as goyalsau demonstrated.

Thanks Rahul and Thanks goyalsau for posting this question.
Doubt cleared !
But one thing, can this problem be solved without using the alternate secant theorem, which goyalsau mentioned, in a GMAT approach.

[rishab1988]

Guys could someone explain why EDC=x?

[beat_gmat_09]

Guys could someone explain why EDC=x?

Do you mean DCE instead of EDC ?
Cause EDC is equal to ADE as depicted in the figure, and is equal to x (as per goyalsau's solution)
The alternate secant theorem can be understood from image below.




In the problem figure applying this theorem Angle EDA equals angle DCE = x.

[rishab1988]

Nope EDC.

Do the double markings indicate that the two angles are equal?

[beat_gmat_09]

Nope EDC.

Do the double markings indicate that the two angles are equal?

Yes.