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by clock60 » Wed May 19, 2010 1:58 pm
colakumarfanta wrote:In triangle ABC , AD is perpendicular to BC , Angle B = 60 degree , Angle A = 45 degree , AB = 5 , value of AC ?
hi
is it possible to give answer choices. i am not sure that got right answer and mine is too ugly
AC=10*3^1/2/(6^1/2+2^1/2)

or after transforming not better
AC=2,5*3^1/2(6^1/2-3^1/2)
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by colakumarfanta » Wed May 19, 2010 9:12 pm
OA is (5 sqrt 6)/2 but not sure
Last edited by colakumarfanta on Wed May 19, 2010 10:52 pm, edited 1 time in total.
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by liferocks » Wed May 19, 2010 9:58 pm
to calculate this we need to know the value of cos15 or sin75..not sure if that comes in the scope of GMAT..or am I missing something?
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by mj78ind » Wed May 19, 2010 11:03 pm
I would tend to agree with "Liferocks" .......

Sin 60 = AD / 5 gives us AD = 5 Sqrt(3)/2

Then Sin 75 = AD / AC

Hence AC = 5 sqrt (3)/ (2*Sin 75)
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by akhpad » Wed May 19, 2010 11:56 pm
colakumarfanta wrote: In triangle ABC , AD is perpendicular to BC , Angle B = 60 degree , Angle A = 45 degree , AB = 5 , value of AC ?
Received PM, asking me to reply.

What is the used of "AD is perpendicular to BC"? Are you missing something in question?

Anyway, Lets assume that your Q is correct as data is sufficient to find AC.

In triangle ABC , Angle B = 60 degree , Angle A = 45 degree , AB = 5 , value of AC?

Angle C = 75 degree

sin75 = sin(45 + 30) = sin45 * cos30 + cos45 * sin30 = (1 + sqrt(3) ) / 2 * sqrt(2)

AC / sin60 = 5 / sin75

AC = 5 * sqrt(6) / (1 + sqrt(3) )
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by akhpad » Thu May 20, 2010 12:19 am
colakumarfanta wrote:OA is (5 sqrt 6)/2 but not sure
in your answer, it is taken 1+sqrt(3) as 2, which is not possible.
1+sqrt(3) = 2.732
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by kstv » Thu May 20, 2010 12:47 am
Image

Triangle ABD is a 60 30 90
AD = √3/2 * 5 BD = 5/2

5/2*√3 < AC < 5

Any answer options ?
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by sanju09 » Thu May 20, 2010 3:04 am
colakumarfanta wrote:In triangle ABC , AD is perpendicular to BC , Angle B = 60 degree , Angle A = 45 degree , AB = 5 , value of AC ?
see here too...

https://www.beatthegmat.com/134-side-of- ... 57706.html
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by frank1 » Thu May 20, 2010 4:36 am
This is NGMAT question ...lol
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by tpr-becky » Thu May 20, 2010 9:27 am
Recieved a PM for this one but not sure I am much help.

Because AD is perpendicular to BC - you have a right triangle formed by ABD - it is also a 30:60:90 becuase angle B is 60. thus, if we know side AB is 5 we know that BD is 5/2 and that AD is 5/2(SQRT3) - becuase of the ration of sides of a 30 60 90.

Sin and cos are outside the range of the GMAT so I am not sure how else you would proceed through this question without answer choices. You can't use pythagorean theorem becuase you only have one side of a right triangle. I would say that on the GMAT you either need more information or answer choices that can be rounded to a correct answers. Sorry I couldn't be more helpful.
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by Stacey Koprince » Thu May 20, 2010 12:13 pm
Received a PM asking me to respond.
This is NGMAT question ...lol
I don't think so. I've never seen it before. I also just searched our database and couldn't find it.

There are multiple problems with this post. The source is not cited. The full text of the problem is not cited. The answer choices are missing. The text of the problem itself has clearly been abbreviated; it is not written in complete sentences.

I would be happy to respond once this has been properly posted. I haven't tried the problem but note that multiple people above have mentioned needing to use trig to solve the problem. If that's the case, then this isn't a GMAT problem (or it isn't a very good GMAT problem) and we shouldn't be studying it (assuming that we want to study GMAT problems... which I think is safe to assume, since we're all on Beat the GMAT!).
Please note: I do not use the Private Messaging system! I will not see any PMs that you send to me!!

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by frank1 » Thu May 20, 2010 8:17 pm
Stacey Koprince wrote:Received a PM asking me to respond.
This is NGMAT question ...lol
I don't think so. I've never seen it before. I also just searched our database and couldn't find it.
well this question was not asked by me.
The asker might be kidding when he asked the question (my assumption) so i replied by kidding...

i wrote NGMAT .....N means Not.....just kidding...

but sorry anyway as it misguided you.

thanks
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by sanju09 » Fri May 21, 2010 12:09 am
Wait a while...

Let me correct myself...

I am rewording the correct question to match my triangle ABC, which has CD perpendicular to AB instead AD perpendicular to BC.

In triangle ABC, CD is perpendicular to AB , Angle A = 60 degree , Angle B = 45 degree, BC = 5, value of AB?


Let's have a triangle ABC in which angle A is 60 degrees, angle B is 45 degrees, and hence angle C is 75 degrees, side BC = 5, and side AB =?

Let's now draw CD perpendicular to AB that meets AB in D.

Now have BD = CD = 5/√2 from the 45˚-45˚-90˚ triangle BCD.

Then have AD = CD/√3 = 5/√6 from the 30˚-60˚-90˚ triangle CAD.

Now, AB = AD + DB = 5/√6 + 5/√2

AB = 5√2 (√3 + 3)/6, in its most wanted rational denominator form.

Sorry! This could be a GMAT problem, no Trigonometry is really required.

My apologies...
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by quantskillsgmat » Fri May 21, 2010 12:43 am
i think this cant be gmat question becauze type of traingle that is formed cant be in gmat syllabus.however application of these question can be asked like length of side bd or abt angles of triangle..however solution is
traingle abd will be 30-90-60 traingle.using 30-90-6- rule one can find sides of triangle abd.as ab is given 5 so bd should be 2.5 and ad shd be 2.5root3.intraingle adc angles will be 15-90-75. and you can find sides as ad is known.but is out of context from gmat point of viewas indepth trigo is nt allowed.
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