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Hi all, Please help

by sachin_yadav » Mon Dec 06, 2010 4:00 am
Hi all,

Please help in the following question

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

(1) BD = 6 sqrt 3
(2) x = 60

Answer is D --> Correct (Each statement ALONE is sufficient to answer the question).
My Answer is B --> incorrect (Statement 2 ALONE is sufficient to answer the question, but statement 1 alone is NOT sufficient).

Not sure why answer is D.
Will be waiting for your reply

Regards
Sachin
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by Rahul@gurome » Mon Dec 06, 2010 4:20 am
Image

Refer to the above image, ∆BOD is a right angled triangle with angles 30-60-90 and CD = 6

Statement 1: BD = 6√3
As in ∆BOD, angle BDO = 30° => OD/BD = cos 30° => OD = BD*cos 30° = (6√3)*(√3/2) = 9
OD = OC + CD = OC + 6 => OC = OD - 6 = 9 - 6 = 3

Again, OB/BD = sin 30° => OB = BD*sin 30° = (6√3)*(1/2) = 3√3

In ∆BOC, (OB)² + (OC)² = (BC)²
Thus, (BC)² = (OB)² + (OC)² = (3√3)² + (3)² = 27 + 9 = 36 => BC = 6

Sufficient.

Statement 2: x = 60
x = angle CBD + angle CDB
=> 60° = angle CBD + 30°
=> angle CBD = 30°

∆BCD is an isosceles triangle => BC = CD = 6

Sufficient.

The correct answer is D.
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by sachin_yadav » Mon Dec 06, 2010 10:26 pm
Rahul@gurome wrote:Image

Refer to the above image, ∆BOD is a right angled triangle with angles 30-60-90 and CD = 6

Statement 1: BD = 6√3
As in ∆BOD, angle BDO = 30° => OD/BD = cos 30° => OD = BD*cos 30° = (6√3)*(√3/2) = 9
OD = OC + CD = OC + 6 => OC = OD - 6 = 9 - 6 = 3

Again, OB/BD = sin 30° => OB = BD*sin 30° = (6√3)*(1/2) = 3√3

In ∆BOC, (OB)² + (OC)² = (BC)²
Thus, (BC)² = (OB)² + (OC)² = (3√3)² + (3)² = 27 + 9 = 36 => BC = 6

Sufficient.

Statement 2: x = 60
x = angle CBD + angle CDB
=> 60° = angle CBD + 30°
=> angle CBD = 30°

∆BCD is an isosceles triangle => BC = CD = 6

Sufficient.

The correct answer is D.
Hi Rahul,

Thank you very much for clearing my doubt after which i tried to solve the question by this method and applied the values of sin 30 and cos 30 and got the answer (because i remember the table), but my only confusion is "why we should divide OD by BD i.e OD/BD and conclude as cos 30, and OB by BD i.e OB/BD and conclude as sin 30 (why not this can be cos 30).

I also used 30 - 60 - 90 triangle strategy and got the answer.

30 60 90
a a'sqrt'3 2a

But how and when we should conclude sin 30 and cos 30.


Thank & Regards
Sachin
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by goyalsau » Mon Dec 06, 2010 11:13 pm
sachin_yadav wrote:
But how and when we should conclude sin 30 and cos 30.
There is a phrase that my teacher taught me in 9th or 10th class.

That is. SOH CAH TOA

SOH = Sin Opposite / Hypotenuse

CAH = Cos Adjacent / Hypotenuse

TOA = Tan Opposite / Adjacent

In every right angle triangle , there is always a doubt that which one is perpendicular and which one is Base, As they both meet to make right angle Triangle.

And the answer is Both can be considered as perpendicular or Base depending upon the Nature of the problem.

Always Remember that whenever we take Sin, Its always Opposite side measurement that we take in the Numerator and Whenever We take Cos its always Adjacent side measurement that we take in the numerator.

if you still have any doubt's Just look at this video it will be great Help.
https://www.youtube.com/watch?v=_S35Ht4imhs,
Saurabh Goyal
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by rishab1988 » Tue Dec 07, 2010 2:41 am
This does NOT look like a real GMAT question for GMAT does NOT test trigonometry.

Correct me if I'm wrong.

Of course there will be geniuses like Rahul,who can solve any quant question out there.
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by Night reader » Tue Dec 07, 2010 3:37 am
sachin_yadav wrote:Hi all,

Please help in the following question

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

(1) BD = 6 sqrt 3
(2) x = 60

Answer is D --> Correct (Each statement ALONE is sufficient to answer the question).
My Answer is B --> incorrect (Statement 2 ALONE is sufficient to answer the question, but statement 1 alone is NOT sufficient).

Not sure why answer is D.
Will be waiting for your reply

Regards
Sachin
Geo concept domination:
2 line segments CD=6, BC-?

st(1) given line segm. BD=6 sqrt(3) => line segm-s BD&CD beginning point common => form angle, \_ BDC=30`(degrees) => explore right triangle ABD => \_ BAC=60` => from line segm. BD=6 sqrt(3) calculate AB (proportion 1:srt [3]:2) => line segm-s AB=6, AD=12, AC=12-6=6, AB=AC => isosceles triangle ABC, angles at base are equal, \_ BAC=60` (note - line segm. AB is drawn mnemonically) \_ABC=60` therefore \_BCA=180`-\_ABC-\_ BAC=60` => triangle ABC is equilateral BC=6

st(2) this statement was explicitly explained in previous solutions => angle proportions (supplementary, complement) we derive line segm-s BC=CD

no trigons ...
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by oddball » Tue Dec 07, 2010 5:01 am
Night reader wrote: Geo concept domination:
2 line segments CD=6, BC-?

st(1) given line segm. BD=6 sqrt(3) => line segm-s BD&CD beginning point common => form angle, \_ BDC=30`(degrees) => explore right triangle ABD => \_ BAC=60` => from line segm. BD=6 sqrt(3) calculate AB (proportion 1:srt [3]:2) => line segm-s AB=6, AD=12, AC=12-6=6, AB=AC => isosceles triangle ABC, angles at base are equal, \_ BAC=60` (note - line segm. AB is drawn mnemonically) \_ABC=60` therefore \_BCA=180`-\_ABC-\_ BAC=60` => triangle ABC is equilateral BC=6

st(2) this statement was explicitly explained in previous solutions => angle proportions (supplementary, complement) we derive line segm-s BC=CD

no trigons ...
How do you know triangle ABD is right-angled? And when you are saying "proportion 1:srt [3]:2", what you are using is a direct application of trigonometry without declaring the terms sin or cos.
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by Night reader » Tue Dec 07, 2010 6:12 am
oddball: agree - I used eyeballing ;(
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by rishab1988 » Tue Dec 07, 2010 6:37 am
So, this is a trigonometry question. This question,therefore,is irrelevant ,according to me,as far as GMAT is considered unless someone shows another way to prove why A is sufficient.
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by sachin_yadav » Tue Dec 07, 2010 7:14 am
rishab1988 wrote:So, this is a trigonometry question. This question,therefore,is irrelevant ,according to me,as far as GMAT is considered unless someone shows another way to prove why A is sufficient.
Hi,

I tried this question with 30 - 60 - 90 triangle strategy and got the answer.

30 60 90
a a'sqrt'3 2a

There are two types of triangle strategy:-

1) 45 - 45 - 90
2) 30 - 60 - 90

With the help of the second strategy we can get the answer, but there is one more doubt and i am not sure about it. As this is a "Data sufficiency question", so we have to solve it with the inputs available, then why are we drawing a line between B and O, and making a right angled triangle OBD or right angled triangle ABD.

Instructors, Experts or anyone please advice on this. If we make OBD or ABD a right angled triangle then we can get the value of BC.

Thanks
Sachin
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by Rahul@gurome » Tue Dec 07, 2010 9:45 am
sachin_yadav wrote: As this is a "Data sufficiency question", so we have to solve it with the inputs available, then why are we drawing a line between B and O, and making a right angled triangle OBD or right angled triangle ABD.

Instructors, Experts or anyone please advice on this. If we make OBD or ABD a right angled triangle then we can get the value of BC.
Good thinking. :)
But solution of geometry problems are often accompanies with such special constructions. It's analogous to assumptions we make for arithmetical or algebraical problems. And one more point to note is that drawing a line between B and O is not a part of analysis of statement 2, we are doing it on the basis of question stem itself.
oddball wrote:How do you know triangle ABD is right-angled?
Note that point A is not a fixed point as length AC is not mentioned. We can take point A as any point on the line CD. Thus there will be a point A on CD for which AB will be perpendicular on BD. Thus ABD will be a right-angled triangle.
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by sachin_yadav » Tue Dec 07, 2010 8:08 pm
Rahul@gurome wrote:
Good thinking. :)
But solution of geometry problems are often accompanies with such special constructions. It's analogous to assumptions we make for arithmetical or algebraical problems. And one more point to note is that drawing a line between B and O is not a part of analysis of statement 2, we are doing it on the basis of question stem itself.

Thank you once again Rahul.
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