OG Quant Supp #117

Never knew the rule either, but the degree measure of the exterior angle of a triangle is equal to the sum of the remote interior angles. Therefore AD and BC have the same length, 6.

(A)

Expect a problem of this level to be the first or second question you see on the actual exam.

I'm sorry, can you go step by step, please. I understand that angle DAB + angle ABD = 2x. Then what is the next step? The OA explanation names angle ABD as y and angle DAB as x, hence x+y=2x. But isn't that just an assumption that angle DAB=x? Accordingly, isn't it just an assumption that triangle ABD=isosceles. Couldn't angle DAB=.5x and angle ABD=1.5x?

You are absolutely correct. I think there is a typo in the book or we are missing a key concept.
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I think the answer is (C).cant be (A) for sure

I guess it's C clearly neither is suficient on its own. However combining the two statement we get:


1. DC = BD, angle DBC = 180-72-72 = 36

ADB = 180-72 (thru exterior angle/complementary angles) = 108

DAB+DBA = 72 (exterior angle)

2. Now, let BC=BD= x.

on to the law of cosines

https://en.wikipedia.org/wiki/Cosine_law

1. CD^2 = x^2 + x^2 - 2*x*x*cos36
2. 6[i.e. AD] = x^2 + AB^2 -2*x*AB*cosABD
3. AB = x^2 + (6+CD)^2 - 2x(6+CD)*cos72

plus according to law of sines:

https://en.wikipedia.org/wiki/Law_of_sines

sin (ABD)/AD = Sin (DAB)/BD, or
sin (ABD) / 6 = sin (180-108-ABD)/x

4. sin(ABD)*x = 6 sin(72-ABD)

we have 4 equations in 4 unknowns: CD (will then substitute it in 3), x - expressed as sin ABD in 4; sin(72-ABD) can be reprsented as sum of the two; AB; and cos ABD that can be expressed thru sinABD (sqrt (1-sin^ABD)

so finally by substituting 1 and 4 in 2 and 3, we will have 2 equations in two unknowns AB and sinABD

That system is potentially solvable (don't ask me to!).
Once it's solved, x can be determined.

I know it sounds complicated (and takes a lot more that 1:45..) but anyway.

there is a typo in this problem - missing info actually in OG

the correct version is there in MGMAT geometry (in DS rephrasing section)
and the missing info is:
angle BAD is marked x degrees.

yeah..only with this "BAD is x degrees" information we can solve the problem. Otherwise, I think the answer is E.

Knowing x=36 degrees, the answer should be D but the GMAT prep showed the answer as A.

hence i assume they dont expect us to use/know the law of cosines?

Discussed here: in-triangle-abc-106669.html

Some OG books (just like the image in the initial post) have a typo missing the info that <BAD=x degrees.

But even knowing that <BAD=x degrees the answer is still A not D. GMAT math is not different from math in general so it can not be that a statement is sufficient but GMAT says it's not, just because some concept is not tested on the GMAT. The fact that trigonometry is not tested on the GMAT means that you won't see a question involving trigonometric functions etc. and also means that every geometry question can be solved without knowing trigonometric properties.

Back to the question:In triangle ABC above, what is the length of side BC?

As <BDC=<BCD then the BD=BC. Also as <ADB=180-2x (exterior angle) and the sum of the angles of a triangle is 180 degrees then in triangle ADB we'll have: x+(180-2x)+<ABD=180 --> <ABD=x. Now, we have that <ABD=x=<DAB so AD=BD --> AD=BD=BC.

Question: BC=?

(1) Line segment AD has length 6 --> AD=BD=BC=6. Sufficient.
(2) x = 36 --> we know only angles which is insufficient to get the length of any line segment.

Answer: A.
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Bunuel

if x - 36 degrees, 2x will be 72 degrees, and we would know 2 angles of the triangle and hence we can derive the third angle. Knowing three angles we could use Law of cosines to get the three sides.
Here is a question from the challenge which is similar to the point im trying to make.
ds-angle-formed-by-diagonal-in-rectangle-57397.html#p884841

True. The idea of finding BC just by statement 2 was too fantastic for me. Thanks for clarification. I don't suppose any law of mathematics can find a definitive value for BC just by knowing the angles 70,70,40 because we can make infinite number of triangles with same angle dimensions and varying BC.

You are wrong. Knowing the angles of a triangle is not enough to get the sides as you can have infinitely many similar triangles withe same angles and different sides.
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Sorry! thanks for correcting me. I had sides and angles switched. Effect of being up from 4:30am. By the way, could you comment on the other thread if the gmat expects you to know Law of cosines.

Again: every GMAT geometry question can be answered without trigonometry.




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