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 E

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

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 E

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,

your explanation confused me, NOT BECAUSE you applied a wrong logic, its because your approach and logic both seems correct to me.....but the OA is D, and my approach to solve this question is also strong and valid...
My Approach
my approach is primarily based on the property of 30,60,90 degree triangle which says that the sides are in the the ratio 1:root3:2 or you can also say the side opp to 30 angle to the side opp to 90 degree angle in a ratio 1:2....and this is a full proof valid property...
Now, i draw a line from C perpendicular to line AD and mark that point E...see the below fig




Triangle CED is a 30-60-90 triangle. Using the side ratios of this special triangle, we know that the hypotenuse is two times the smallest leg. Therefore, segment ED is equal to 1.
From this we see that triangle EDB is an isosceles triangle, since it has two equal sides (of length 1). We know that EDB = 120°; therefore angles �DEB and �DBE are both 30°. So an.EBA = 15, also we can conclude that an.AEB = 150, thus in tr. AEB, we get an. EAB = 15
Now notice two other isosceles triangles:
(1) Triangle CEB is an isosceles triangle, since it has two equal angles (each 30 degrees). Therefore segment CE = segment EB.
(2) Triangle AEB is an isosceles triangle, because an. EAB = an.EBA = 15...proved above already, Thus, AE = EB which means as AE = CE as proved earlier CE = EB
So, we can say, Tr.AEC is isosceles
since CEA is 90 degrees, angles ACE and EAC must be equal to 45 degrees each. Therefore angle x = 45 + 30 = 75 degrees. The correct answer is D.

@NR

If u find any flaw in my approach or in yours please share, otherwise we will call some expert

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

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

@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???

sure, I'll wander around math forum and see if the experts shed some light

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???

Here's the erroneous statement:

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.

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).

Stuart Kovinsky wrote:Here's the erroneous statement:

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.

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).

Hi Stuart,
Thanks for reply but still the doubt is not clear. I know the limitations of the property you stated that's why Before arriving on the conclusion that NR's approach is VALID, 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 u can suggest any single shape in which satisfy the given condition (AD divide BC ratio 2:1) and angleCAD NOT EQUALS 30 degree, that would be really helpful.

Thanks and Regards

Anurag@Gurome wrote:

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.

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.

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.

Thanks Anurag for your useful analysis
I think in free time i should draw the actual triangle with the help of "D"

Thanks again