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In triangle ABC, if angle ABC is 30 degrees, AC = 2*sqrt(2) and AB = B

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In triangle ABC, if angle ABC is 30 degrees, AC = 2*sqrt(2) and AB = B  [#permalink]

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New post 10 Feb 2016, 11:20
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In triangle ABC, if angle ABC is 30 degrees, \(AC = 2*\sqrt 2\) and AB = BC = X, what is the value of X?

(A) \(\sqrt 3 – 1\)

(B) \(\sqrt 3 + 2\)

(C) \(\frac{(\sqrt 3 – 1)}{2}\)

(D) \(\frac{(\sqrt 3 + 1)}{2}\)

(E) \(2*(\sqrt 3 + 1)\)

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Re: In triangle ABC, if angle ABC is 30 degrees, AC = 2*sqrt(2) and AB = B  [#permalink]

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New post 12 Feb 2016, 07:55
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Bunuel wrote:
In triangle ABC, if angle ABC is 30 degrees, \(AC = 2*\sqrt 2\) and AB = BC = X, what is the value of X?

(A) \(\sqrt 3 – 1\)

(B) \(\sqrt 3 + 2\)

(C) \(\frac{(\sqrt 3 – 1)}{2}\)

(D) \(\frac{(\sqrt 3 + 1)}{2}\)

(E) \(2*(\sqrt 3 + 1)\)



Hi,
Just saw some discussion on this Q, I too will put in my bit to it..
Most of the Qs can be answered by elimination and standard methods...

1) POE


So A straight forward way to eliminate all wrong choices will be
Image
Take B as center of a circle and draw arc AC..
now Arc AC > line AC..
arc AC>2\(\sqrt{2}\)..
arc AC= circumference of circle * 30/360= 2*pi*x*30/360= 2*pi*x/12..
so 2*pi*x/12>2\(\sqrt{2}\)...
x> 2*7*12*2\(\sqrt{2}\)/(2*22)..
x> 84\(\sqrt{2}\)/22...
this is nearly equal to 4\(\sqrt{2}\) ..
so x is nearly equal to 5.6

lets see the choices..
(A) \(\sqrt{3}-1 \approx 0.73\)

(B) \(\sqrt{3}+2 \approx 3.73\)

(C) \(\frac{\sqrt{3}-1}{2} \approx 0.36\)

(D) \(\frac{\sqrt{3}+1}{2} \approx 1.36\)

(E) \(2(\sqrt{3}+1) \approx 5.46\)
clearly E is the answer


2)algebric way


Although rarely tested, If I see a Q on this line, where an angle and opposite sides are given, apply cosine rule..
This triangle would mean two sides of x with third side 2\(\sqrt{2}\) and opposite angle 30..
\(AC^2=BC^2+AB^2-2*BC*AB*cos 30\)...
\({(2\sqrt{2})}^2=x^2+x^2-2*x*x*\sqrt{3}/2\)
\(8=2x^2-x^2\sqrt{3}\)..
\(x^2=8/(2-\sqrt{3})\)..
we can simplify
E is the answer
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In triangle ABC, if angle ABC is 30 degrees, AC = 2*sqrt(2) and AB = B  [#permalink]

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New post Updated on: 08 Jul 2016, 11:44
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There are two approaches to this problem - we can solve it directly, or we can examine the answer choices and come to a conclusion about which one must be correct.

Algebraic approach:
Draw a perpendicular from point A to side BC, label it point D. Draw another perpendicular from point B to side AC, label it point E. Now we have a 30-60-90 triangle and a two similar 15-75-90 triangles. Let the length of CD = a
Attachment:
Isosceles Triangle.png
Isosceles Triangle.png [ 4.68 KiB | Viewed 28282 times ]


Now we can use some ratios to figure out x

Looking at triangle ABD, we know that \(\frac{BD}{AB} = \frac{x-a}{x} = \frac{\sqrt{3}}{2}\) (1)

And looking at triangles ADC and BEC (similar triangles) we can see that \(\frac{AC}{DC} = \frac{BC}{EC} = \frac{2\sqrt{2}}{a}=\frac{x}{\sqrt{2}}\) (2)

From (2) \(ax = 4\), or \(a=\frac{4}{x}\)

Cross multiply (1) and plug in \(a=\frac{4}{x}\)
\(2x-2a = \sqrt{3}x\)

\(2x-2(\frac{4}{x})=\sqrt{3}x\)

\(2x-\frac{8}{x}-\sqrt{3}=0\) --> multiply by \(x\)

\(2x^2-8-\sqrt{3}x^2=0\) --> gather terms

\((2-\sqrt{3})x^2=8\)

\(x=\sqrt{\frac{8}{2-\sqrt{3}}}\) --> Now this doesn't look like any of our answer choices, so we will have to manipulate it a bit

\(x=\frac{2\sqrt{2}}{\sqrt{2-\sqrt{3}}}\) --> multiply top and bottom by the conjugate of the denominator to rationalize the denominator

\(x=\frac{2\sqrt{2}*\sqrt{2+\sqrt{3}}}{(\sqrt{2-\sqrt{3}})*(\sqrt{2+\sqrt{3}})}\)

\(x=\frac{2\sqrt{2}*\sqrt{2+\sqrt{3}}}{1}\) --> bring the \(\sqrt{2}\) into the square root

\(x=2\sqrt{4+2\sqrt{3}}\) --> Now rearrange the expression to complete the square within the square root

\(x=2\sqrt{1+2\sqrt{3}+3} = 2\sqrt{(1+\sqrt{3})^2}\)

\(x=2(1+\sqrt{3})\)

Answer: E


WHEW! Ok, now the simpler approach that involves almost no calculation, just a few quick estimates and a good idea of what's going on in the triangle:

Because \(\angle{B}\) is \(30^{\circ}\), and the side opposite \(\angle{B}\) is \(2\sqrt{2}\), we know that the other two sides, x, will be longer than \(2\sqrt{2}\).

\(2\sqrt{2}\) is \(\approx 2.8\)

Now if we look at the answer choices:
(A) \(\sqrt{3}-1 \approx 0.73\)

(B) \(\sqrt{3}+2 \approx 3.73\)

(C) \(\frac{\sqrt{3}-1}{2} \approx 0.36\)

(D) \(\frac{\sqrt{3}+1}{2} \approx 1.36\)

(E) \(2(\sqrt{3}+1) \approx 5.46\)

Immediately we can eliminate A, C and D. To choose between B and E, we need to ask whether x should be almost twice as much as AC, or less than 1.5*AC. Go back to the diagram and look at triangle ABD. Since it is a 30-60-90 triangle, we know that AB (i.e. \(x\)) = 2*AD. Now, though the diagram is not to scale, it is accurate enough to surmise that AD is only slightly less than AC. Therefore, \(x\) should be only slightly less than 2*AC, which matches with answer E.

I recommend the second approach, but it only works well because the answer choices are spread out enough. That being said, I don't think the GMAT will expect you to do the algebraic approach, and will design the answer choices specifically to be able to use approximation.

Cheers
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Originally posted by davedekoos on 11 Feb 2016, 12:44.
Last edited by davedekoos on 08 Jul 2016, 11:44, edited 1 time in total.
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Re: In triangle ABC, if angle ABC is 30 degrees, AC = 2*sqrt(2) and AB = B  [#permalink]

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New post 10 Feb 2016, 11:45
1
answer is E.

we know that the sides are same so the angle opposite the side will also be the same.
Angle will be 75.
Applying the formula.

sina/a=sinb/b.we can get the answer sin 75=root6+root2/2
Hope it is correct
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Re: In triangle ABC, if angle ABC is 30 degrees, AC = 2*sqrt(2) and AB = B  [#permalink]

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New post 10 Feb 2016, 17:10
1
we see here is an isosceles triangle with one angle as 30 degrees and other two angles as (180 – 30)/2 = 75 degrees each.

The side opposite the 30 degrees angle is 2*sqrt(2). One simple observation is that X must be greater than 2*sqrt(2) because these sides are opposite the greater angles (75 degrees).

2*sqrt(2) is a bit less than 2*1.5 because Sqrt(2) = 1.414. So 2*sqrt(2) is a bit less than 3. Note that options (A), (C), and (D) are much smaller than 3, so these cannot be the value of X. ONLY 2 CHOICES LEFT B OR E . AFTER DOING CALLCULATION AND KNOWING THE FACT THAT ,A greater side of a triangle is opposite a greater angle. MY ANSWR IS E
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In triangle ABC, if angle ABC is 30 degrees, AC = 2*sqrt(2) and AB = B  [#permalink]

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New post 12 Feb 2016, 06:41
1
First, I would like to thank ‘davedekoos’ for the graph. I will use the same graph as above.

Second, I will try to solve using estimations at end

It is advisable to know to know √2 & √3 which is 1.4 & 1.7 respectively. So 2√2= 2*1.4=2.8
Coupled with the above the fact, the larger side opposite to largest angle. Checking the answers quickly

A) 1.7- 1= 0.7 < 2.8………. out (should be larger)

B) 1.7+2= 3.7 possible keep

C) 0.7/2 out

D) 2.7/2 out

E) 2*2.7= 5.4 possible keep

The two possible answers are spread out so more likely to proceed with estimation
Going back to the graph above

In right triangle ABD (30-60-90) which is powerful
AD= (1/2) AB = (1/2)X ………….. since it is opposite to Angle 30
BD= (√3/2) AB= (√3/2)X


In right triangle ADC
AD = X/2
AC= 2√2
DC= BC- BD= X – X√3/2 = X (1- √3/2) = 0.15X…….. I will try to estimate higher to 0.25X to use the fraction ¼
(AC)^2= (AD)^2+ (DC)^2
(2√2)^2= (X/2)^2 + (X/4)^2
8 = (X^2)/4 + (X^2)/16 …….multiply by 16

16*8= 4 X^2 + X^2 =
16*8=5 X^2………..> since we make coefficient of X^2 higher when we used 1/4 o we can decrease to 4 to make division easier

16 *8 =4 X^2..............> X^2= 32 ........> 5<X<6 so we sure that our target Answer is E

The power of estimation is evident in this question.
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Re: In triangle ABC, if angle ABC is 30 degrees, AC = 2*sqrt(2) and AB = B  [#permalink]

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New post 08 Jul 2016, 11:26
chetan2u wrote:
Bunuel wrote:
In triangle ABC, if angle ABC is 30 degrees, \(AC = 2*\sqrt 2\) and AB = BC = X, what is the value of X?

(A) \(\sqrt 3 – 1\)

(B) \(\sqrt 3 + 2\)

(C) \(\frac{(\sqrt 3 – 1)}{2}\)

(D) \(\frac{(\sqrt 3 + 1)}{2}\)

(E) \(2*(\sqrt 3 + 1)\)



Hi,
Just saw some discussion on this Q, I too will put in my bit to it..
Most of the Qs can be answered by elimination and standard methods...

1) POE


So A straight forward way to eliminate all wrong choices will be
Image
Take B as center of a circle and draw arc AC..
now Arc AC > line AC..
arc AC>2\(\sqrt{2}\)..
arc AC= circumference of circle * 30/360= 2*pi*x*30/360= 2*pi*x/12..
so 2*pi*x/12>2\(\sqrt{2}\)...
x> 2*7*12*2\(\sqrt{2}\)/(2*22)..
x> 84\(\sqrt{2}\)/22...
this is nearly equal to 4\(\sqrt{2}\) ..
so x is nearly equal to 5.6

lets see the choices..
(A) \(\sqrt{3}-1 \approx 0.73\)

(B) \(\sqrt{3}+2 \approx 3.73\)

(C) \(\frac{\sqrt{3}-1}{2} \approx 0.36\)

(D) \(\frac{\sqrt{3}+1}{2} \approx 1.36\)

(E) \(2(\sqrt{3}+1) \approx 5.46\)
clearly E is the answer



I tried the approximation method to very similar question and it not worked
in-the-diagram-what-is-the-value-of-x-129962.html
:cry:
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Re: In triangle ABC, if angle ABC is 30 degrees, AC = 2*sqrt(2) and AB = B  [#permalink]

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New post 09 Jul 2016, 02:20
I have used approximation:

In triangle ABC,angle B=30 degrees and AB=AC=X,which means it is an isosceles triangle.
Now drop a perpendicular from point A on BC.Let us call this point D
We get a 30-60-90 triangle-ABD and another triangle which is 15-75-90 triangle-ADC.

Now AB=X
then AB/BD=2/sq root 3(In 30-60-90 triangle,the sides are in the ratio 1:sq rt3:2 opposite to angle 30-60-90)
So,BD=sq rt3*X/2

So,DC=2-sqrt 3/2*X

Similarly AB/AD=2/1
X/AD=2/1
or AD=X/2

Now we have only one variable X,so we need one equation to solve it.
Using pythagoras theorem for triangle ADC,
(2*sq rt2)^2=(2-sqrt3/2*x)^2+(X/2)^2

solving,we get X^2=8/2-sq rt 3
Rationalising we get x^2=8(2+sq rt 3)

sq rt 3=1.732,
2+1.732=3.732*8=29.856(somewhere between sq root of 25 and 36)
so X=5. xxx

Check the answer choices.E is the correct option
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Re: In triangle ABC, if angle ABC is 30 degrees, AC = 2*sqrt(2) and AB = B  [#permalink]

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New post 26 Jul 2017, 11:31
chetan2u wrote:
Bunuel wrote:
In triangle ABC, if angle ABC is 30 degrees, \(AC = 2*\sqrt 2\) and AB = BC = X, what is the value of X?

(A) \(\sqrt 3 – 1\)

(B) \(\sqrt 3 + 2\)

(C) \(\frac{(\sqrt 3 – 1)}{2}\)

(D) \(\frac{(\sqrt 3 + 1)}{2}\)

(E) \(2*(\sqrt 3 + 1)\)



Hi,
Just saw some discussion on this Q, I too will put in my bit to it..
Most of the Qs can be answered by elimination and standard methods...

1) POE


So A straight forward way to eliminate all wrong choices will be
Image
Take B as center of a circle and draw arc AC..
now Arc AC > line AC..
arc AC>2\(\sqrt{2}\)..
arc AC= circumference of circle * 30/360= 2*pi*x*30/360= 2*pi*x/12..
so 2*pi*x/12>2\(\sqrt{2}\)...
x> 2*7*12*2\(\sqrt{2}\)/(2*22)..
x> 84\(\sqrt{2}\)/22...
this is nearly equal to 4\(\sqrt{2}\) ..
so x is nearly equal to 5.6

lets see the choices..
(A) \(\sqrt{3}-1 \approx 0.73\)

(B) \(\sqrt{3}+2 \approx 3.73\)

(C) \(\frac{\sqrt{3}-1}{2} \approx 0.36\)

(D) \(\frac{\sqrt{3}+1}{2} \approx 1.36\)

(E) \(2(\sqrt{3}+1) \approx 5.46\)
clearly E is the answer


2)algebric way


Although rarely tested, If I see a Q on this line, where an angle and opposite sides are given, apply cosine rule..
This triangle would mean two sides of x with third side 2\(\sqrt{2}\) and opposite angle 30..
\(AC^2=BC^2+AB^2-2*BC*AB*cos 30\)...
\({(2\sqrt{2})}^2=x^2+x^2-2*x*x*\sqrt{3}/2\)
\(8=2x^2-x^2\sqrt{3}\)..
\(x^2=8/(2-\sqrt{3})\)..
we can simplify
E is the answer


Thanks a lot for the short cut method.
Can you please provide any link where similar kind of tips are summarized [specially for Geometry] ?
It will help a lot.
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Re: In triangle ABC, if angle ABC is 30 degrees, AC = 2*sqrt(2) and AB = B  [#permalink]

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New post 29 Sep 2018, 23:45
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