Problem Solving

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 ?

OA is 38

goyalsau wrote:OA is 38

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

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

goyalsau wrote: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.

beat_gmat_09 wrote: 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,

goyalsau wrote: 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.

beat_gmat_09 wrote:

goyalsau wrote: 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 wrote:

beat_gmat_09 wrote:

goyalsau wrote: 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.

Guys could someone explain why EDC=x?

rishab1988 wrote: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.

Nope EDC.

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

rishab1988 wrote:Nope EDC.

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

Yes.