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atulmangal wrote:In the figure, point D divides side BC of triangle ABC into segments BD and DC of lengths 1 and 2 units respectively. Given that angleADC = 60º and angleABD = 45º, what is the measure of angle x in degrees? (Note:Figure is not drawn to scale.)
(A) 55 (B) 60 (C) 70 (D) 75 (E) 90
@NR,Night reader wrote:supplementary angle ADB=(180-60)=120; angle DAB=(180-120-45)=15. Given the opposite angle/side proportion we can deduce that if angle DAB=15 and its opposing side is 1, then angle CAD opposing to the side 2 will be (15*2)=30. Angle CAB=(30+15)=45 and angle ACD or x=180-(45+45)=90
IOM Eatulmangal wrote:In the figure, point D divides side BC of triangle ABC into segments BD and DC of lengths 1 and 2 units respectively. Given that angleADC = 60º and angleABD = 45º, what is the measure of angle x in degrees? (Note:Figure is not drawn to scale.)
(A) 55 (B) 60 (C) 70 (D) 75 (E) 90
atulmangal wrote:@NR
If u find any flaw in my approach or in yours please share, otherwise we will call some expert
@NR,Night reader wrote:good, this problem which has been devised as a challenge geometry question and was originally posted on GMAT club web-site https://gmatclub.com/forum/solution-chal ... -2998.html would be tricky.
I read your explanation which is also replicated with the solution of Manhattan GMAT - we operate with 30-60-90 triangles and handle the isosceles triangle properties. It's good overview, and I actually calculated sides to prove that answer D geometrically is possible. So this is a clean solution and the answer should be D, after several minutes of solution though.
p.s. ROI from such problems is high only if one is aiming Q50(51).
atulmangal wrote:@NR
If u find any flaw in my approach or in yours please share, otherwise we will call some expert
atulmangal wrote: @NR,
But still the point is, the approach u stated earlier, though conclude wrong answer, is still seems VALID theoretically in this context as both the TR.CAD and TR.ABD are sharing the same base...and a shorter one...so whats wrong out there....???
I'm interested to know about the flaw in your approach because if that approach clicked me rather than the approach i followed, in that case i;m surely gonna solve the question through your approach as its a shorter one...and in that case m gonna get a wrong answer, and i'm targeting a perfect score in quant because last time without much prep i scored 49 IN QUANT, so just wanna clear this confusion...thats why i asked you about the flaw?? can we call some expert to resolve this issue???
For circles, double the interior angle does mean double the opposing arc length, but the same rule doesn't hold true for triangles unless you're dividing up a right angle (and since angle CAB isn't right, it won't be a simple 1:2 ratio).we can deduce that if angle DAB=15 and its opposing side is 1, then angle CAD opposing to the side 2 will be (15*2)=30.
Hi Stuart,Stuart Kovinsky wrote:Here's the erroneous statement:
For circles, double the interior angle does mean double the opposing arc length, but the same rule doesn't hold true for triangles unless you're dividing up a right angle (and since angle CAB isn't right, it won't be a simple 1:2 ratio).we can deduce that if angle DAB=15 and its opposing side is 1, then angle CAD opposing to the side 2 will be (15*2)=30.
Yes, your logic is true that all the length of the line segments and measure of angles are fixed in the figure provided. But the measure of angle CAD is not equal to 30 degrees, it is equal to 45 degrees and that comes directly from your detailed analysis of the problem.atulmangal wrote:...i actually tried on paper to get any possible shape, in which Line AD divide Line BC in the ratio 2:1 WITHOUT making an angle of 30 degree with line AC (i;e the double of angle DAB), BUT i FAILED. The line BC is rigid, and line AD is dividing the line BC in 2:1 ratio, again a stated fact which has to be true and if you try to draw on paper then you only get
angle CAD =30 degree....NO OTHER CONSTRUCTION OR FIGURE POSSIBLE.
Thanks Anurag for your useful analysisAnurag@Gurome wrote:Yes, your logic is true that all the length of the line segments and measure of angles are fixed in the figure provided. But the measure of angle CAD is not equal to 30 degrees, it is equal to 45 degrees and that comes directly from your detailed analysis of the problem.atulmangal wrote:...i actually tried on paper to get any possible shape, in which Line AD divide Line BC in the ratio 2:1 WITHOUT making an angle of 30 degree with line AC (i;e the double of angle DAB), BUT i FAILED. The line BC is rigid, and line AD is dividing the line BC in 2:1 ratio, again a stated fact which has to be true and if you try to draw on paper then you only get
angle CAD =30 degree....NO OTHER CONSTRUCTION OR FIGURE POSSIBLE.
If the measure of angle CAD has been 30 degrees, then the figure is not possible. Let's assume the measure of angle CAD is 30 degrees and see what happens!
If angle CAD = 30° => angle CAB = (30 + 15)° = 45°
Hence ABC is an isosceles right-angled triangle with angle ACB = 90° and AC = BC = (2 + 1) = 3
Now, in right-angled triangle ACD, CD = 2 and AC = 3 => AD = √13
But, ratio of the lengths of these line segments do not match the ratio for a 30-60-90 triangle.
Hence, the figure is not possible!
Now, try the same with angle CAD = 45°, there will be no such violation.
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