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Senior Manager  V
Joined: 25 Sep 2018
Posts: 430
Location: United States (CA)
Concentration: Finance, Strategy
GMAT 1: 640 Q47 V30 GPA: 3.97
WE: Investment Banking (Investment Banking)
If AB < BC < CA, AD = DC, and each angle of triangle ACB is a whole  [#permalink]

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Difficulty:   65% (hard)

Question Stats: 55% (03:00) correct 45% (02:43) wrong based on 34 sessions

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Attachment: GC2.png [ 14.76 KiB | Viewed 424 times ]
If AB < BC < CA, AD = DC, and each angle of triangle ACB is a whole number of degrees, which of the following is the maximum possible value of angle DAB? (Note: Diagram is not drawn to scale)
A) 9°
B) 14°
C) 19°
D) 21°
E) 51°

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Director  G
Joined: 19 Oct 2013
Posts: 516
Location: Kuwait
GPA: 3.2
WE: Engineering (Real Estate)
Re: If AB < BC < CA, AD = DC, and each angle of triangle ACB is a whole  [#permalink]

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Abhi077 wrote:
Attachment:
GC2.png
If AB < BC < CA, AD = DC, and each angle of triangle ACB is a whole number of degrees, which of the following is the maximum possible value of angle DAB? (Note: Diagram is not drawn to scale)
A) 9°
B) 14°
C) 19°
D) 21°
E) 51°

Given that CA > BC > AB.

This depends on understanding that the longest side corresponds to the largest angle.

Let DAB = x

we know the angle opposing AB is 50 + 1 = 51

We also know that CAB = 50 + x

The sum of angles = 180

51 + 50 + x + CBA = 180

x + CBA = 79

Try x = 19

then CBA = 60, however, CAB = 69 incorrect.

Try x = 14

then CBA = 65 and CAB = 64. Correct

Manager  P
Joined: 01 Mar 2015
Posts: 69
If AB < BC < CA, AD = DC, and each angle of triangle ACB is a whole  [#permalink]

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Abhi077 wrote:
Attachment:
GC2.png
If AB < BC < CA, AD = DC, and each angle of triangle ACB is a whole number of degrees, which of the following is the maximum possible value of angle DAB? (Note: Diagram is not drawn to scale)
A) 9°
B) 14°
C) 19°
D) 21°
E) 51°

We have AB < BC < CA
In a triangle angle opposite to largest side has highest value and the one corresponding to smallest side has lowest value, therefore

Angle ACB < Angle BAC < Angle CBA
=> 51° < 50°+x < Angle CBA
(using sum of angles of triangle = 180°)
=> 51° < 50°+x < 180°- 51°- (50°-x)
=> 51° < 50°+x < 79°- x

To find highest value of x,
50°+x < 79°- x
=> 2x < 79°- 50°
=> x < 29°/2
=> x < 14.5°
As all angles are whole numbers, highest value of x is 14°

Posted from my mobile device If AB < BC < CA, AD = DC, and each angle of triangle ACB is a whole   [#permalink] 18 Oct 2018, 12:08
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