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Math Expert V
Joined: 02 Sep 2009
Posts: 58257
In triangle ABC, if angle ABC is 30 degrees, AC = 2*sqrt(2) and AB = B  [#permalink]

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Difficulty:   65% (hard)

Question Stats: 60% (03:06) correct 40% (03:04) wrong based on 119 sessions

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

_________________
Math Expert V
Joined: 02 Aug 2009
Posts: 7952
Re: In triangle ABC, if angle ABC is 30 degrees, AC = 2*sqrt(2) and AB = B  [#permalink]

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3
2
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 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
Attachments IMG_5544.JPG [ 1.79 MiB | Viewed 48619 times ]

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Manager  Joined: 09 Jul 2013
Posts: 107
In triangle ABC, if angle ABC is 30 degrees, AC = 2*sqrt(2) and AB = B  [#permalink]

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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 [ 4.68 KiB | Viewed 33440 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

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

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

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
_________________
Dave de Koos

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.
General Discussion
Intern  B
Joined: 13 Apr 2015
Posts: 31
Re: In triangle ABC, if angle ABC is 30 degrees, AC = 2*sqrt(2) and AB = B  [#permalink]

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1

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
Intern  Joined: 10 Feb 2016
Posts: 5
WE: Sales (Consulting)
Re: In triangle ABC, if angle ABC is 30 degrees, AC = 2*sqrt(2) and AB = B  [#permalink]

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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
SVP  V
Joined: 26 Mar 2013
Posts: 2341
In triangle ABC, if angle ABC is 30 degrees, AC = 2*sqrt(2) and AB = B  [#permalink]

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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
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
(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.
VP  D
Joined: 05 Mar 2015
Posts: 1015
Re: In triangle ABC, if angle ABC is 30 degrees, AC = 2*sqrt(2) and AB = B  [#permalink]

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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 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 Intern  Joined: 28 Dec 2015
Posts: 37
Re: In triangle ABC, if angle ABC is 30 degrees, AC = 2*sqrt(2) and AB = B  [#permalink]

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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

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
Manager  B
Joined: 14 Jun 2016
Posts: 70
Location: India
GMAT 1: 610 Q49 V21 WE: Engineering (Manufacturing)
Re: In triangle ABC, if angle ABC is 30 degrees, AC = 2*sqrt(2) and AB = B  [#permalink]

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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 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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Posts: 13082
Re: In triangle ABC, if angle ABC is 30 degrees, AC = 2*sqrt(2) and AB = B  [#permalink]

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_________________ Re: In triangle ABC, if angle ABC is 30 degrees, AC = 2*sqrt(2) and AB = B   [#permalink] 29 Sep 2018, 23:45
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