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In the attached figure, point D divides side BC of triangle [#permalink]
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15 Oct 2003, 00:57
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This topic is locked. If you want to discuss this question please repost it in the respective forum. In the attached figure, point D divides side BC of triangle ABC into segments BD and DC of lengths 1 and 2 units respectively. Given angle ADC = 60 deg and angle ABD = 45, what is angle x? Please explain how you got your answer. (Figure is not drawn to scale.)
(A) 55
(B) 60
(C) 70
(D) 75
(E) 90
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AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993
Last edited by AkamaiBrah on 15 Oct 2003, 22:38, edited 2 times in total.



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Re: Challenge Geometry Triangles [#permalink]
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15 Oct 2003, 02:14
AkamaiBrah wrote: In the attached figure, point D divides side BC of triangle ABC into segments BD and DC of lengths 1 and 2 units respectively. Given angle ADC = 60 deg and angle ABD = 45, what is angle x? Please explain how you got your answer. (Figure is not drawn to scale.) (A) 55 (B) 60 (C) 70 (D) 75 (E) 90
I think its D.
i wish i had time to explain this.
thanks
praetorian



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brain torture again... Akamaibrah, is it possible to solve this problem without employing theorem of sines?
My long way: angle DAB=15 deg. and angle ADB=120. DB=1. So, in triangle ADB we know all the angles and one side. Therefore, we can calculate via the theorem of sines all the other elements including side AD. Since CD=2, in triangle ACD we know two sides and angle CDA=60 deg. and so may calculate X using the above theorem.
Honestly, up to now, I see no other methods.



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stolyar wrote: brain torture again... Akamaibrah, is it possible to solve this problem without employing theorem of sines?
My long way: angle DAB=15 deg. and angle ADB=120. DB=1. So, in triangle ADB we know all the angles and one side. Therefore, we can calculate via the theorem of sines all the other elements including side AD. Since CD=2, in triangle ACD we know two sides and angle CDA=60 deg. and so may calculate X using the above theorem.
Honestly, up to now, I see no other methods.
It is possible to solve this without using any trigonometry or other fancy math whatsoever. After all, this is a GMAT forum. (BTW, drawing the figure carefully, then using a protractor to measure x is not an acceptable GMAT technique).
Use the force, Luke....
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How About (E) 90?
logic:
CDA is 60 so ADB is 120
This makes DAB = 15  Side with length 1 is opposite this angle (1)
Side CD is 2 so the angle opposite is must be 30  from (1)
Making CDA a right 306090 triangle
Correct?



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Re: My Shot [#permalink]
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15 Oct 2003, 19:37
exy18 wrote: How About (E) 90?
logic: CDA is 60 so ADB is 120 This makes DAB = 15  Side with length 1 is opposite this angle (1) Side CD is 2 so the angle opposite is must be 30  from (1) Making CDA a right 306090 triangle
Correct?
Doubling the angle does not double the length of the opposite side.
Hint: there is a 60 degree angle here. What shapes have 60 degree angles and what special properties of such shapes can we use to add more information to the problem?
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AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993



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ans is b. 60
Exterior angle ADB is equal to the sum of interior opposite angles(
DAC + ACD).
120(ADB) = x+ angle CAD
since, angle CAD = 15+45 ==>60 (sum of interior opposite angles for triangle
BAD)
X = 180  angle ADB+Angle CAD ===> 180 (60+60) = 60



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araspai wrote: ans is b. 60 Exterior angle ADB is equal to the sum of interior opposite angles( DAC + ACD). 120(ADB) = x+ angle CAD since, angle CAD = 15+45 ==>60 (sum of interior opposite angles for triangle BAD) X = 180  angle ADB+Angle CAD ===> 180 (60+60) = 60
Sorry, you cannot conclude that CAD = 15 + 45 = 60. Those two angles are not interior opposite angles. Moreover, an "exterior" angle to a triangle must have one of its sides common with the triangle.
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Solution:
The answer is (D).
As first, it appears that there is not enough information to compute the rest of the angles in the diagram. especially since you are given only one angle and one side in triangle ACD and no obvious way of getting any more information short of actually measuring something or using trigonometry tricks.
In a geometry problem, when faced with situations such as this, you should look for ways to add construction lines to exploit any special properties of the given diagram and expose "concealed information."
For example, note than the figure contains a 60 deg angle, and two lines with lengths of ratio of 2 to 1. How can we use this information? Remember that a 306090 righttriangle also has a ratio of 2 to 1 for the ratio of its hypotenuse to its short leg. This suggests that by dropping a construction line from C to line AD and forming a right triangle we identify another segment of length 1 and perhaps have a foothold into deducing more information about the diagram.
LetтАЩs drop a line from C to point E to form a right triangle, then connect points E and B as follows:
From the properties of a 306090 triangle, we can set segment ED equal to 1. You will see that the addition of this simple bit of knowledge opens the floodgates to complete information about the diagram!!!
We can now note that triangle EDB is an isosceles triangle. angle EDB = 180 тАУ 60 = 120;, hence angles DEB and EBD are both 30;. Since angle ECD is also 30, triangle CEB is isosceles and segment CE = EB. Angle AEB = 180 тАУ 30 = 150, angle ABE = 45 тАУ 30 = 15, and angle BAE = 180 тАУ (15 + 150) = 15. This means than triangle AEB is isosceles and AE = EB. Since CE is also equal to EB, AE = CE and triangle ACE is isosceles with angle CAE = angle ACE = 45; (recall that angle DEC = angle AEC = 90). Hence, x = angle ACE + angle ECD = 45 + 30 = 75 and the answer is (D).
Hope you liked this problem. Big Kudos for everyone who made the attempt  i hope you learned something from it!
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AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993



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Re: In the attached figure, point D divides side BC of triangle [#permalink]
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Re: In the attached figure, point D divides side BC of triangle
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