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A point Q is located within the interior of a square ABCD so that it
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01 May 2015, 01:48
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Re: A point Q is located within the interior of a square ABCD so that it
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10 May 2015, 08:24
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 . Answer 135. attached image.
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A point Q is located within the interior of a square ABCD so that it
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01 May 2015, 11:41
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 \(1+42*1*2*cos(a) = 5  4*cos(a) = X^2\) (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
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02 May 2015, 13:51
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
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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
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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



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A point Q is located within the interior of a square ABCD so that it
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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 \(1+42*1*2*cos(a) = 5  4*cos(a) = X^2\) (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 . thanks.
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Re: A point Q is located within the interior of a square ABCD so that it
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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.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
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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.



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Re: A point Q is located within the interior of a square ABCD so that it
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15 Jan 2017, 16:29
Hi Bunuel, can you please post the OA to this question?
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Re: A point Q is located within the interior of a square ABCD so that it
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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. Bunuel, Could you provide the official solution to this problem? It is pretty difficult to solve. Thanks.
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