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01 May 2015, 01:48
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D
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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
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10 May 2015, 08:24
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Bunuel
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.
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General Discussion
01 May 2015, 11:41
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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
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
