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goyalsau: Legendary Member; Posts: 866; Joined: Mon Aug 02, 2010 6:46 pm; Location: Gwalior, India; Thanked: 31 times

by goyalsau » Tue Nov 23, 2010 4:23 am

Saurabh Goyal
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Source: Beat The GMAT — Problem Solving | Tue Nov 23, 2010 4:23 am

beat_gmat_09: Master | Next Rank: 500 Posts; Posts: 437; Joined: Sat Nov 22, 2008 5:06 amFollowed by: 1 member; Location: India; Thanked: 50 times; GMAT Score:580

by beat_gmat_09 » Tue Nov 23, 2010 6:48 am

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 ?

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goyalsau: Legendary Member; Posts: 866; Joined: Mon Aug 02, 2010 6:46 pm; Location: Gwalior, India; Thanked: 31 times

by goyalsau » Tue Nov 23, 2010 7:37 am

OA is 38

Saurabh Goyal
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diebeatsthegmat: Legendary Member; Posts: 1119; Joined: Fri May 07, 2010 8:50 amFollowed by: 3 members; Thanked: 29 times

by diebeatsthegmat » Wed Nov 24, 2010 1:20 am

goyalsau wrote:OA is 38

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

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goyalsau: Legendary Member; Posts: 866; Joined: Mon Aug 02, 2010 6:46 pm; Location: Gwalior, India; Thanked: 31 times

by goyalsau » Wed Nov 24, 2010 1:42 am

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

Saurabh Goyal
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beat_gmat_09: Master | Next Rank: 500 Posts; Posts: 437; Joined: Sat Nov 22, 2008 5:06 amFollowed by: 1 member; Location: India; Thanked: 50 times; GMAT Score:580

by beat_gmat_09 » Wed Nov 24, 2010 3:21 am

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.

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goyalsau: Legendary Member; Posts: 866; Joined: Mon Aug 02, 2010 6:46 pm; Location: Gwalior, India; Thanked: 31 times

by goyalsau » Wed Nov 24, 2010 5:03 am

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,

Saurabh Goyal
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beat_gmat_09: Master | Next Rank: 500 Posts; Posts: 437; Joined: Sat Nov 22, 2008 5:06 amFollowed by: 1 member; Location: India; Thanked: 50 times; GMAT Score:580

by beat_gmat_09 » Wed Nov 24, 2010 5:07 am

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.

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goyalsau: Legendary Member; Posts: 866; Joined: Mon Aug 02, 2010 6:46 pm; Location: Gwalior, India; Thanked: 31 times

by goyalsau » Wed Nov 24, 2010 5:18 am

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

Saurabh Goyal
[email protected]
-------------------------

EveryBody Wants to Win But Nobody wants to prepare for Win.

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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

by Rahul@gurome » Wed Nov 24, 2010 5:42 am

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 90Â° 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 Î¸ = 38Â° and method is as goyalsau demonstrated.

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beat_gmat_09: Master | Next Rank: 500 Posts; Posts: 437; Joined: Sat Nov 22, 2008 5:06 amFollowed by: 1 member; Location: India; Thanked: 50 times; GMAT Score:580

by beat_gmat_09 » Wed Nov 24, 2010 5:54 am

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 90Â° 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 Î¸ = 38Â° 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.

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rishab1988: Master | Next Rank: 500 Posts; Posts: 332; Joined: Tue Feb 09, 2010 3:50 pmFollowed by: 7 members; Thanked: 41 times; GMAT Score:720

by rishab1988 » Wed Nov 24, 2010 6:36 am

Guys could someone explain why EDC=x?

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beat_gmat_09: Master | Next Rank: 500 Posts; Posts: 437; Joined: Sat Nov 22, 2008 5:06 amFollowed by: 1 member; Location: India; Thanked: 50 times; GMAT Score:580

by beat_gmat_09 » Wed Nov 24, 2010 9:59 am

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.

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rishab1988: Master | Next Rank: 500 Posts; Posts: 332; Joined: Tue Feb 09, 2010 3:50 pmFollowed by: 7 members; Thanked: 41 times; GMAT Score:720

by rishab1988 » Wed Nov 24, 2010 10:07 am

Nope EDC.

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

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beat_gmat_09: Master | Next Rank: 500 Posts; Posts: 437; Joined: Sat Nov 22, 2008 5:06 amFollowed by: 1 member; Location: India; Thanked: 50 times; GMAT Score:580

by beat_gmat_09 » Wed Nov 24, 2010 10:12 am

rishab1988 wrote:Nope EDC.

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

Yes.

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