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tejal777
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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1) Is insufficient - This length doesn't help us except to tell us that BCD is not a right isosceles triangle.
2) x = 60. This means that Angle BCD = 180-60 =120. The remaining angle CBD = 180-120-30 = 30.
This tells us that the triangle is isosceles. BC = CD = 6.

ANS = B
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Nope..that's what I thought too!!please try again..
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Nope..that's what I thought too!!please try again..

If CD = 6, what is the length of BC?

As it's a DS question no need to actually find the value of BC, rather than to determine that it's possible to find it with either of statements:

(1) \(BD=6\sqrt{3}\). We know CD, BD and the angle between them. The opposite side BC is fixed and has single value, meaning that you cannot draw two or more triangles with given two sides and the angle between them. Sufficient.

(2) \(x=60\). Again we know x, hence we know all the angles in triangle BCD, plus we know one of the sides CD=6, again only one such triangle exists, hence the length of BC can be determined. Sufficient.

Answer: D.
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Kudos to both of you. Had not thought about that.

So just for my knowledge, if we know 2 sides of a triangle, and the angle in between we can safely determine that the information is sufficient?

When answering I did think about what you guys said, but thought that we a) couldn't assume that if we have 2 fixed sides and an angle we can derive the 3rd (althgouh I know we don't need to actaully derive it) and b) that trigonometry wasn't really required knowledge for the GMAT?
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Bunuel's way is absolutly correct!!
And no trig is not reqd:)
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Yup. Bunuel is bang on...And yes, trig knowledge is NOT required for GMAT, but if you have a basic understanding, it helps.
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Yup. Bunuel is bang on...And yes, trig knowledge is NOT required for GMAT, but if you have a basic understanding, it helps.

wow..very good tip..never knew that...

would a question like this actually appear on an exam?

if so, I got some studying to do
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From statement 1 we can know that the triangle BDC is 30-60-90 degree because cd=6 and bd =6root3 so for a 30-60-90 triangle x-xroot3-2x=6-6root3-12 so the length of BC =12
From Statement 2 I dont have asolution yet
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:cry:
Using Premise 1) when BD is given,BC is given ,use cosine formula
BC ^ 2 = BD ^ 2 + + CB ^ 2 - 2 CB.BC Cos 30
2) When x = 60, BCD=120 , Triangle is isoceles , CB=6
therefore D, either premise satisfies
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Attachment:
The attachment g2.JPG is no longer available
If CD = 6, what is the length of BC?

(1) \(BD=6\sqrt{3}\)
(2) x = 60

One way to solve this would be to draw a perpendicular line from B to A to make a 90 deg triangle and then solve.

Statement 1: \(BD=6\sqrt{3}\) - Sufficient

For Triangle BAD we know <C = 30, <A = 90, so <B = 60. We now have a 30:60:90 triangle with sides in the ratio x:x\(\sqrt{3}\):2x. Knowing BD allows us to calculate the value of x = 3\(\sqrt{3}\). So AD = 9 and hence AC = 3 and BA = 3\(\sqrt{3}\).

Now for triangle BAC we know the 2 sides we can calculate BC the 3rd side. Sufficient.

Statement 2: x = 60 - Sufficient
If x = 60 then <C = 120 and <B = 30. We now have an isosceles triangle with two equal sides making BC and CD equal (corresponding angles being equal) so BC = 6. Sufficient.
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If CD = 6, what is the length of BC?

(1) BD=6\sqrt{3}

We know the length of BD and CD. We also know the angle that lies between them. BC is fixed. We don't have to find the solution, just confirm that there is only one possible solution. The only way that BC's angle or length could change is if line DC were extended to the left but as the diagram states, that isn't possible because we are given it's length. Sufficient.

(2) x = 60

If x = 60 then the interior angle c = 120. Seeing as angle d = 30, angle b also = 60. The shape and side lengths of this triangle are fixed into position and cannot be changed. Given side length CD = 6 then there is only one possible answer for BC. Sufficient.
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(1) we know that CD = 6, and angle CDB = 30
we can draw a perpendicular line, and get 2 right triangles (30-60-90), since the new angle CFD must be 90 degrees, we can conclude that angle C must be 60 degrees. Thus, we have CD = 6, CF = 3, and FD = 3 sqrt 3. Hm, this is interesting, 3 sqrt 3 is half of BD. That means that CF is the median of BD. Knowing FD & BF, we can calculate for BC. but that is not needed. Statement 1 Sufficient

(2) x = 60, thus we can conclude that we have an isosceles triangle, and BC = CD.
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Hi bunnel pls correct me if my approach is wrong.
Here angle D is 30

In statement 1 BD 6root3

Draw a perpendicular from Cto BD at O then angle at O 90 angle D 30 angle C 60.
If CD =6 then side opposite to D 30 degree is half then CO is 3 then OD is 3root3
then OB= 3root3.
Trianlge OCD are congruent OCB then BC =CD. Value of BC=6
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Draw BE perpendicular to AD, let \(CE = y\).

stmt1:
\(sin30 = BE/BD\), BD is known, so BE can be found as sin30 =1/2.
\(cos30 = DE/BD\), \(\sqrt{3}/2 = (6+y)/BD\) , so y can be found.
Now we know BE and CE, apply pythagoras and find BC.

stmt2:
\(sin30 = BE/BD\), BD is known, so BE can be found as sin30 =1/2.
Now apply \(sin60 = BE/BC\), so BC can be found.

D

Hey

How do we apply Pythagoras while we don't know the angle of ECB?

Thanks
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