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

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

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New post 01 May 2015, 01:48
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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
trig1c.gif [ 2.81 KiB | Viewed 3142 times ]

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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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New post 10 May 2015, 08:24
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Bunuel wrote:
Image
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 .

Answer 135.
attached image.
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gmatclub.jpg [ 67.42 KiB | Viewed 2434 times ]


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

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New post 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
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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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New post 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
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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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New post 04 May 2015, 04:04
could you please help me find out how you managed to get the diagonal as 4?

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
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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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New post 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
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A point Q is located within the interior of a square ABCD so that it  [#permalink]

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New post 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



I hope that this http://gmatclub.com/forum/download/file.php?id=26605 helps .

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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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New post 10 Jul 2015, 23:15
Lucky2783 wrote:
Bunuel wrote:
Image
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 .

Answer 135.
attached image.


Can you Pls. tell me how we got PD = 1?
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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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New post 11 Jul 2015, 02:13
Bunuel, can you post your official solution, please? TIA.

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

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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New post 16 Jan 2017, 01:13
Bunuel wrote:
Image
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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