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# A point Q is located within the interior of a square ABCD so that it

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Math Expert
Joined: 02 Sep 2009
Posts: 50001
A point Q is located within the interior of a square ABCD so that it  [#permalink]

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01 May 2015, 01:48
1
11
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Difficulty:

95% (hard)

Question Stats:

29% (01:48) correct 71% (02:25) wrong based on 114 sessions

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A point Q is located within the interior of a square ABCD so that it is 1 unit from vertex A, 2 units from vertex B, and 3 units from vertex D. The measure of angle AQB is:

A. between 90º and 120º
B. 120º
C. between 120º and 135º
D. 135º
E. greater than 135º

Kudos for a correct solution.

Attachment:

trig1c.gif [ 2.81 KiB | Viewed 3142 times ]

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Joined: 07 Aug 2011
Posts: 540
GMAT 1: 630 Q49 V27
Re: A point Q is located within the interior of a square ABCD so that it  [#permalink]

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10 May 2015, 08:24
1
6
Bunuel wrote:

A point Q is located within the interior of a square ABCD so that it is 1 unit from vertex A, 2 units from vertex B, and 3 units from vertex D. The measure of angle AQB is:

A. between 90º and 120º
B. 120º
C. between 120º and 135º
D. 135º
E. greater than 135º

Kudos for a correct solution.

Attachment:
The attachment trig1c.gif is no longer available

those who have used trigonometry , please note that none of the GMAT question requires us to know sin cosine and tan formulas, knowing them is just a luxury not a necessity .

attached image.
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##### General Discussion
Manager
Joined: 17 Mar 2015
Posts: 116
A point Q is located within the interior of a square ABCD so that it  [#permalink]

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01 May 2015, 11:41
1
Thought about the adequate approach for like half an hour but gave up, Here is the solution but I wonder what is the adequate way (the GMAT way)

Let the side of the square be X then the diagonal will be X*$$\sqrt{2}$$
angle AQB = a
angle BQD = b
$$1+4-2*1*2*cos(a) = 5 - 4*cos(a) = X^2$$ (1)
$$4 + 9 - 2*2*3*cos(b) = 13 - 12*cos(b) = X^2$$ (2)
$$1 + 9 - 2*1*3*cos(360-(a+b)) = 10 - 6*cos(a+b) = 2*X^2$$ (3)
(1) and (2)
$$5 - 4*cos(a) = 13 - 12*cos(b) => cos(b) = (8 + 4*cos(a)/12 = (cos(a) + 2)/3$$
(1) and (3)
$$5 - 4*cos(a) = 5 - 3*cos(a+b)$$
$$3*cos(a+b) = 4*cos(a)$$
$$3*cos(a)*cos(b) - 3*sin(a)*sin(b) = 4*cos(a)$$
$$cos(a)*(cos(a)+2) - 3*sin(a)*\sqrt{1 - cos^2(b)} = 4*cos(a)$$
reduce the cos(a)
$$cos(a)+2 - 3*tg(a)*\sqrt{1 - (cos^2(a) + 4*cos(a) + 4)/9} = 4$$
$$tg(a)*\sqrt{5 - cos^2(a) - 4*cos(a)} = 2 - cos(a)$$
$$tg^2(a)*(5 - cos^2(a) - 4*cos(a) = 4 - 4*cos(a) + cos^2(a)$$
$$5*tg^2(a) - sin^2(a) - 4*tg^2(a)*cos(a) = 4 - 4*cos(a) + cos^2(a)$$
$$5*tg^2(a) - 4*tg^2(a)*cos(a) = 5 - 4*cos(a)$$
$$tg^2(a)*(5 - 4*cos(a)) = 5 - 4*cos(a)$$
$$tg^2(a) = 1$$
$$tg(a) = 1$$ or $$tg(a) = -1$$
$$a = 45$$ or $$a = 135$$

D
Manager
Joined: 01 Jan 2015
Posts: 56
Re: A point Q is located within the interior of a square ABCD so that it  [#permalink]

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02 May 2015, 13:51
1
D - 135
AD diag = 4; diag of sq - root2a=4 ; a = 2root2
Now we have 3 sides as 1,3,2root2 and 1^2+2root2sr=9 which is 3^2 hence angle = 90 +half of it 45 = 135
Intern
Joined: 06 Apr 2015
Posts: 35
Re: A point Q is located within the interior of a square ABCD so that it  [#permalink]

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04 May 2015, 04:04

If and only if angle AQD = 180 degrees will diagonal be equal to 4.

Ted21 wrote:
D - 135
AD diag = 4; diag of sq - root2a=4 ; a = 2root2
Now we have 3 sides as 1,3,2root2 and 1^2+2root2sr=9 which is 3^2 hence angle = 90 +half of it 45 = 135
Intern
Joined: 10 Jul 2014
Posts: 13
Re: A point Q is located within the interior of a square ABCD so that it  [#permalink]

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10 May 2015, 01:19
Ted21 wrote:
D - 135
AD diag = 4; diag of sq - root2a=4 ; a = 2root2
Now we have 3 sides as 1,3,2root2 and 1^2+2root2sr=9 which is 3^2 hence angle = 90 +half of it 45 = 135

angle to be measured is aqb whose 2 sides are 1,2 not 1,3
Director
Joined: 07 Aug 2011
Posts: 540
GMAT 1: 630 Q49 V27
A point Q is located within the interior of a square ABCD so that it  [#permalink]

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10 May 2015, 08:33
Zhenek wrote:
Thought about the adequate approach for like half an hour but gave up, Here is the solution but I wonder what is the adequate way (the GMAT way)

Let the side of the square be X then the diagonal will be X*$$\sqrt{2}$$
angle AQB = a
angle BQD = b
$$1+4-2*1*2*cos(a) = 5 - 4*cos(a) = X^2$$ (1)
$$4 + 9 - 2*2*3*cos(b) = 13 - 12*cos(b) = X^2$$ (2)
$$1 + 9 - 2*1*3*cos(360-(a+b)) = 10 - 6*cos(a+b) = 2*X^2$$ (3)
(1) and (2)
$$5 - 4*cos(a) = 13 - 12*cos(b) => cos(b) = (8 + 4*cos(a)/12 = (cos(a) + 2)/3$$
(1) and (3)
$$5 - 4*cos(a) = 5 - 3*cos(a+b)$$
$$3*cos(a+b) = 4*cos(a)$$
$$3*cos(a)*cos(b) - 3*sin(a)*sin(b) = 4*cos(a)$$
$$cos(a)*(cos(a)+2) - 3*sin(a)*\sqrt{1 - cos^2(b)} = 4*cos(a)$$
reduce the cos(a)
$$cos(a)+2 - 3*tg(a)*\sqrt{1 - (cos^2(a) + 4*cos(a) + 4)/9} = 4$$
$$tg(a)*\sqrt{5 - cos^2(a) - 4*cos(a)} = 2 - cos(a)$$
$$tg^2(a)*(5 - cos^2(a) - 4*cos(a) = 4 - 4*cos(a) + cos^2(a)$$
$$5*tg^2(a) - sin^2(a) - 4*tg^2(a)*cos(a) = 4 - 4*cos(a) + cos^2(a)$$
$$5*tg^2(a) - 4*tg^2(a)*cos(a) = 5 - 4*cos(a)$$
$$tg^2(a)*(5 - 4*cos(a)) = 5 - 4*cos(a)$$
$$tg^2(a) = 1$$
$$tg(a) = 1$$ or $$tg(a) = -1$$
$$a = 45$$ or $$a = 135$$

D

thanks.
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Intern
Joined: 30 Jul 2008
Posts: 19
Re: A point Q is located within the interior of a square ABCD so that it  [#permalink]

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10 Jul 2015, 23:15
Lucky2783 wrote:
Bunuel wrote:

A point Q is located within the interior of a square ABCD so that it is 1 unit from vertex A, 2 units from vertex B, and 3 units from vertex D. The measure of angle AQB is:

A. between 90º and 120º
B. 120º
C. between 120º and 135º
D. 135º
E. greater than 135º

Kudos for a correct solution.

Attachment:
trig1c.gif

those who have used trigonometry , please note that none of the GMAT question requires us to know sin cosine and tan formulas, knowing them is just a luxury not a necessity .

attached image.

Can you Pls. tell me how we got PD = 1?
Current Student
Joined: 25 Jan 2015
Posts: 91
Location: India
Concentration: Strategy, Sustainability
Re: A point Q is located within the interior of a square ABCD so that it  [#permalink]

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11 Jul 2015, 02:13

Bunuel wrote:
A point Q is located within the interior of a square ABCD so that it is 1 unit from vertex A, 2 units from vertex B, and 3 units from vertex D. The measure of angle AQB is:

A. between 90º and 120º
B. 120º
C. between 120º and 135º
D. 135º
E. greater than 135º

Kudos for a correct solution.

Attachment:
trig1c.gif
Current Student
Status: DONE!
Joined: 05 Sep 2016
Posts: 384
Re: A point Q is located within the interior of a square ABCD so that it  [#permalink]

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15 Jan 2017, 16:29
Hi Bunuel, can you please post the OA to this question?

Thanks!!!
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Joined: 14 Nov 2016
Posts: 1314
Location: Malaysia
Re: A point Q is located within the interior of a square ABCD so that it  [#permalink]

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16 Jan 2017, 01:13
Bunuel wrote:

A point Q is located within the interior of a square ABCD so that it is 1 unit from vertex A, 2 units from vertex B, and 3 units from vertex D. The measure of angle AQB is:

A. between 90º and 120º
B. 120º
C. between 120º and 135º
D. 135º
E. greater than 135º

Kudos for a correct solution.

Attachment:
trig1c.gif

Bunuel, Could you provide the official solution to this problem? It is pretty difficult to solve. Thanks.
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Re: A point Q is located within the interior of a square ABCD so that it  [#permalink]

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18 Apr 2018, 21:21
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