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What is the area of a triangle ABC in which angle BAC = 35

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Intern
Joined: 06 Jul 2009
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What is the area of a triangle ABC in which angle BAC = 35 [#permalink]

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06 Jul 2009, 17:10
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What is the area of a triangle ABC in which angle BAC = 35 degrees?
1. Angle ACB = 60 degrees
2. AB = 7, BC = 5
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Joined: 20 Jun 2009
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Re: Geometry DS [#permalink]

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07 Jul 2009, 01:49
Tough question - this one requires knowledge of trigonometry beyond the basic formula for the area of a triangle. To find the area of a non-right triangle, we need one of the following sets of information:

1. Two sides and an angle opposite one of the known sides
2. Two angles and any side
3. Two sides and their included angle
4. All three sides

As this is a data sufficieny question, the actual calculations don't matter - we simply need to see whether either statement provides us with sufficient information to satisfy any of 1-4.

Statement 1 is insufficient. We can determine that angle ABC must equal 85 degrees (and hence know the measure of each angle), but cannot calculate area without knowing anything about the sides of the triangle.

Statement 2 is sufficient. Per number (1) above, we can determine the area of a triangle if we know the length of two sides and an angle opposite one of the sides. We know from the stem that BAC = 35 degrees; from statement 2, we also know that the length of AC is 7 and BC (the side opposite a known angle) is 5. Thus, we have all the requisite information to calculate the area of the triangle. Choice B is correct.

I'm curious to know the source of this question. I was under the impression that the GMAT wouldn't test knowledge of trigonometry beyond the basics (e.g., the ratios of the sides of 45-45-90 and 30-60-90 triangles, the pythagorean theorem and the basic area formula).
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Re: Geometry DS [#permalink]

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07 Jul 2009, 02:53
B is suff. from trigonometry. Area is (7.cos 35 + 5 root(1- (7/5.sin35)^2)).7sin35/2 (area is from trigonometry and not important).
A is not suff. because we can have so many triangles that have the angles (35-60-85)
Manager
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Re: Geometry DS [#permalink]

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09 Jul 2009, 23:11
Simple question.
Answer B.
It is kinda difficult to explain without drawing a figure. I tried drawing in a word document but am not able to.

The concept it that from point B draw a line to join AC which cuts AC at 90 degree. This is our altitude.
Leave it..............too difficult to explain.Sorry.
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Joined: 05 Jul 2008
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Re: Geometry DS [#permalink]

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10 Jul 2009, 03:35
RockyBalboa wrote:
What is the area of a triangle ABC in which angle BAC = 35 degrees?
1. Angle ACB = 60 degrees
2. AB = 7, BC = 5

C for me.
1. clearly insuf
2.There are 2 distinct triangle which satisfy all.
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Re: Geometry DS [#permalink]

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10 Jul 2009, 11:53
Thanks for the figure DavidArchuleta.
Now I can try to explain this.

Stmt 1 - Not suff. All agree

Stmt 2 - Suff.
We know Angle A = 35
AB = 7
BC2 = 5
Angle AC1B = 90 (means BC1 is height of traingle with base as AC2)

Now in right traingle AC1B we have -
BC1/BA = SIN(35).................(Sin X = Perpendicular / Hypotanuse)
So we know BC1.
BC1/AC1 = TAN(35)...............(TAN X = Base / Hypotanuse)

So we know height of traingle = BC1
We need to find the base of trianle or we need to know AC2.

AC2 = AC1 + C1C2
We already calculated AC1, we need to calculate C1C2 now.

Again in right triangle BC1C2 we use Pythagoras theorem -
to get C1C2.

We have the base now and also the altitude.
Area can be calculated.
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Re: Geometry DS [#permalink]

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11 Jul 2009, 04:03
mdfrahim wrote:
Thanks for the figure DavidArchuleta.
Now I can try to explain this.

Stmt 1 - Not suff. All agree

Stmt 2 - Suff.
We know Angle A = 35
AB = 7
BC2 = 5
Angle AC1B = 90 (means BC1 is height of traingle with base as AC2)

Now in right traingle AC1B we have -
BC1/BA = SIN(35).................(Sin X = Perpendicular / Hypotanuse)
So we know BC1.
BC1/AC1 = TAN(35)...............(TAN X = Base / Hypotanuse)

So we know height of traingle = BC1
We need to find the base of trianle or we need to know AC2.

AC2 = AC1 + C1C2
We already calculated AC1, we need to calculate C1C2 now.

Again in right triangle BC1C2 we use Pythagoras theorem -
to get C1C2.

We have the base now and also the altitude.
Area can be calculated.

I have no idea what you are talking about :D. What I said is just there are 2 trianlges satisfying that they have BAC= 35, AB=7, BC=5 so it is insufficient to determine the only value of the area of triangle ABC. sin35=0.57 so BCA cann't be 90 so there are 2 points C existing if we already have A and B.
I confused between answer E and C because the triangle which has BAC=35, ABC=60, AB=7, BC=5 may not exist. But it doesn't matter.
C for me!
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Re: Geometry DS [#permalink]

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11 Jul 2009, 17:15
Read the expalantion again by modifying the figure to represent angle BC1A as 90.
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Re: Geometry DS [#permalink]

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12 Jul 2009, 00:46
mdfrahim wrote:
Read the expalantion again by modifying the figure to represent angle BC1A as 90.

Why? ) Why is BC1A=90?
Ok, I get cha!
You don't understand what I mean.
The question asks whether or not we can determine THE ONLY VALUE of the area of the triangle ABC. There are 2 triangles satisfying all the conditions so there are 2 values of the area.
Somebody help me! I'm not a native English speaker!
http://74.125.153.132/search?q=cache:nA ... vi&ct=clnk
The value based data sufficiency question stem means one and only one value for x. A statement that gives a range or two or more values for x is insufficient!
If there is only one value that satisfies each statement, then together the statements are sufficient.
If there is still more than one possible value, then the statements alone and combined are insufficient.

http://www.platinumgmat.com/about_gmat/ds_techniques i've just found this
Statements Producing Two Values Are Not Automatically Insufficient
This is probably one of the dirtiest tricks the GMAT test-writers can pull on a medium-level test-taker. Just because a statement, when simplified, yields two values does not mean the statement is automatically insufficient. Those two values could produce the same value for the question, in which case the statement is sufficient.

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Last edited by DavidArchuleta on 12 Jul 2009, 02:04, edited 3 times in total.
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Re: Geometry DS [#permalink]

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12 Jul 2009, 00:56
carriedinterest wrote:
Tough question - this one requires knowledge of trigonometry beyond the basic formula for the area of a triangle. To find the area of a non-right triangle, we need one of the following sets of information:

1. Two sides and an angle opposite one of the known sides
2. Two angles and any side
3. Two sides and their included angle
4. All three sides

Statement 2 is sufficient. Per number (1) above, we can determine the area of a triangle if we know the length of two sides and an angle opposite one of the sides. We know from the stem that BAC = 35 degrees; from statement 2, we also know that the length of AC is 7 and BC (the side opposite a known angle) is 5. Thus, we have all the requisite information to calculate the area of the triangle. Choice B is correct.

1. Two sides and an angle opposite one of the known sides
This statement is not right.
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Re: Geometry DS [#permalink]

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12 Jul 2009, 09:47
DavidArchuleta , Do you agree that area of triangle = 1/2 * base * height
If using the given info. of stmt 2 we can somehow calculate base and height then it will be enough........agree or disagree ?
Now in your initial figure can we assume Base = AC2...............agree or disagree ?
If AC2 is the base then what will be the height ? It will be nothing but a line connecting B and AC2 which is perpendicular on AC2.....agree or disagree ?
So the height is BC1 which makes angle BC1A as a right triangle...............agree or disagree.

If you agree with everything written above then only look at the below calculations.
We will first attemp to calculate the height (BC1).
Angle AC1B is a right triangle with angle A as 35 and AB = 7
Using trigonometry,
BC1/AB = Sin(35)
BC1 = 7*Sin(35)
BC1 = 7 * 0.57358
BC1 = 4.01
So the height is 4.01..............agree or disagree ?

Now we will try to calculate the base.
Base is AC2 which = AC1 + C1C2
Lets dertmine C1C2 first using right triangle BC1C2
Using hypotanuse theorem,
P^2 + B^2 = H^2
BC1^2 + C1C2^2 = BC2^2
(4.01)^2 + C1C2^2 = 5^2
C1C2^2 = 25 - 16.08
C1C2 = 2.97

Now calculate AC1 using triangle AC1B using trigonometry ,
AC1/AB = Cos(35)
AC1 = 7 * cos35
Ac1 = 7*0.81915
AC1 = 5.73

So base = 2.97 + 5.73 = 8.7

You are able to determine the base and height both. Hence suff.

Note - There was no need for all these calculations. This is only for demo. purposes. While solving this question I did not calculated anything however just determined what can be calculated. This question took me less then 45 seconds to solve.
No matter how you draw the triangle the area will always come out to be same.
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Re: Geometry DS [#permalink]

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12 Jul 2009, 23:04
I'm done.
In my figure, BC1 is not the height of triangle ABC2. BH is. BH is the height of the 2 triangles ABC1 and ABC2. In my figure, AB=7, BC1=BC2=5. (again, BC1 or BC2 is not perpendicular to AC2, BH is perpendicular to AC2)
Your calculation is ok, but it's not neccessary. I totally understand what youre saying. The problem is that you don't get the question. Is there another value for the triangle ABC? Yes, it is. There are 2 distinct triangles so there are 2 values. Your calculation is one of them. So B is not sufficient!

Quote:
Now we will try to calculate the base.
Base is AC2 which = AC1 + C1C2
Lets dertmine C1C2 first using right triangle BC1C2

Can AC2= AC1-C1C1 happen? Who says that C1 lies between A and C2 =))
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Re: Geometry DS [#permalink]

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12 Jul 2009, 23:33
Try to calculate the other value of the area.I will be interested to know that
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Re: Geometry DS [#permalink]

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13 Jul 2009, 00:26
mdfrahim wrote:
Try to calculate the other value of the area.I will be interested to know that

Haha, can I use yours?
Angle AC1B is a right triangle with angle A as 35 and AB = 7
Using trigonometry,
BC1/AB = Sin(35)
BC1 = 7*Sin(35)
BC1 = 7 * 0.57358
BC1 = 4.01
So the height is 4.01
Now we will try to calculate the base.
Base is AC2 which = AC1 - C1C2
Lets dertmine C1C2 first using right triangle BC1C2
Using hypotanuse theorem,
P^2 + B^2 = H^2
BC1^2 + C1C2^2 = BC2^2
(4.01)^2 + C1C2^2 = 5^2
C1C2^2 = 25 - 16.08
C1C2 = 2.97

Now calculate AC1 using triangle AC1B using trigonometry ,
AC1/AB = Cos(35)
AC1 = 7 * cos35
Ac1 = 7*0.81915
AC1 = 5.73

AC2= AC1-C1C2=5.73-2.97=2.76
I will send this link to my friend, who is gonna study Phd. in Maths this year. He will say I'm stupid haha.
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Re: Geometry DS [#permalink]

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13 Jul 2009, 05:46
Why don't we settle this once and for all - OA, please!
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Re: Geometry DS [#permalink]

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21 Jul 2009, 07:34
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I think David's explanation makes the answer C pretty crystal clear. I don't feel the need to wait for the OA. I think the trouble you are having with his answer is reconciling the facts with what you have learned in trig. What needs to be clarified is your trig - not the answer to this question.
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Re: Geometry DS [#permalink]

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03 Aug 2009, 15:08
i am biased towards C as well,

can we say that a triangle is identified by lenght of 2 sides and the inclosed angle not any other angel ...am i right??
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Re: Geometry DS [#permalink]

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03 Aug 2009, 15:50
need both statements to be sufficient,

if only know statement 1, then we only know the form of the triangle, the size could vary
if only know statement 2, area could change based on what degree angle B is
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Re: Geometry DS [#permalink]

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03 Aug 2009, 16:25
yezz wrote:
i am biased towards C as well,

can we say that a triangle is identified by lenght of 2 sides and the inclosed angle not any other angel ...am i right??

right, if the known angle is enclosed in between the two known sides, then we have a identified triangle here.
Re: Geometry DS   [#permalink] 03 Aug 2009, 16:25
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