Data Sufficiency

vikram4689 wrote:This will help. c^2 = a^2 + b^2 - 2ab cos(C)

You don't need the cosine law to solve this.
In fact, you don't need any knowledge of sine, cosine and tangent on the GMAT.

Cheers,
Brent

gnod wrote:ok so obviously i'm having some trouble understanding solutions.



i don't understand how statement A is sufficient.
The video solution states that since we're given that BD = 6 sq. rt 3, that makes triangle BCD only one specific triangle with side 6, 6 sq. rt 3 and BC.
But isn't this asking what specifically the value of BC is?

Here are 2 ways to see that statement 1 is sufficient.

Big point: For geometry DS questions, we are typically checking to see whether the statements "lock" a particular angle or length into having just one value.

Given statement 1, we see that there can be only one triangle BCD since we know the lengths of sides BD and CD and we know the angle between them. This information "locks" side BC into having only 1 length.
What is the length? It doesn't matter.
Since our information locks the length of side BC, we must have enough info, which means statement 1 is sufficient.

Or . . . here's another way to look at statement 1:


Cheers,
Brent

A lot of upper-level geometry questions require us to add extra lines here and there. Sometimes those extra lines help and sometimes they don't.
For example, when I first approached this question (I had forgotten the question entirely since it has been over 2 years since I created it), I tried to create a different right triangle by drawing a line from point B that is perpendicular to AC. I didn't see how this line would help me, so I abandoned it and tried adding another line, and that one helped.

I mention all of this because mathematical solutions often make it look like the author just stepped up to a question and immediately chose the best way to tackle it. If students believe that this is how questions are typically solved, they become concerned if this is not their experience with math.

The truth is that mathematics (especially geometry) is often performed in the dark. First you try one path of inquiry and attempt to gauge whether that path is bringing you closer to the solution (which is not always easy to tell). If it's not bringing you closer, you abandon it for another approach. Then another. And so on.

The more solutions you see, the more tools you collect for your mathematical toolkit. With a larger toolkit, you have even more paths to choose from when you tackle the next tricky math question.

Cheers,
Brent

subhakam wrote: Brent - i am not very clear on the concept of the altitude. Per my understanding , altitude when drawn from the vertex of an angle to the unequal side in an ISOSCELES triangle divides the side (to which the altitude/perpendicular bisector) in to 2 equal parts. Here how can we assume both sides are the same? Unless we take the value of B (x=60) and see that both the angles are 30? Hence isosceles? Please help me understand if such assumptions can be made?

I wasn't assuming that the triangle is an isosceles triangle when I concluded that BE has length 3root3.

When I drew that perpendicular line, I created a 30-60-90 "special" right triangle.
Since one side (CD) has length 6, I was able to find the other two lengths.
Most importantly, I found that side ED has length 3root3
IMPORTANT: I found all of these measurements without using statement 1.

When I get to statement 1 (BD = 6root3), I know that side BE must have length 3root3 (since I already know that side ED has length 3root3)
So, that's how I concluded that side BE must have length 3root3

Cheers,
Brent