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# In the figure, point D divides side BC of triangle ABC into segments

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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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12 Nov 2016, 20:57
enigma123 wrote:

In the figure, point D divides side BC of triangle ABC into segments BD and DC of lengths 1 and 2 units respectively. Given that ÐADC = 60º and ÐABD = 45º, what is the measure of angle x in degrees? (Note:
Figure is not drawn to scale.)

A. 55
B. 60
C. 70
D. 75
E. 90

Any idea guys what will be the correct answer please? Also any idea how can I cut and paste the pictures in my post? Is it possible?

Attachment:
The attachment Triangle.png is no longer available

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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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15 Nov 2016, 22:36
Hi Bunuel
please clarify how we decide that triangle CAO is isoseles and AO and line AO is s.root of 3
Also how we decide that line OB is s.root of 3
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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Updated on: 14 Feb 2017, 11:32
Bunuel wrote:
enigma123 wrote:

In the figure, point D divides side BC of triangle ABC into segments BD and DC of lengths 1 and 2 units respectively. Given that ÐADC = 60º and ÐABD = 45º, what is the measure of angle x in degrees? (Note:
Figure is not drawn to scale.)

(A) 55
(B) 60
(C) 70
(D) 75
(E) 90

Any idea guys what will be the correct answer please? Also any idea how can I cut and paste the pictures in my post? Is it possible?

Complete solution for all the angles is in the image below:

x=45+30=75.

Notes:
Sides with one blue segment crossing them are equal and sides with two blues segments crossing them are equal too.

CO is perpendicular to AD --> OD=1 (from 30-60-90 right triangle property as sides are in ratio $$1:\sqrt{3}:2$$) --> as OD=BD=1 then ODB is an isosceles triangle.

<CDO and <BDO are supplementary to each other (supplementary angles are two angles that add up to 180°), so <BDO=120 --> <DAB=180-(120+45)=15.

As ODB is an isosceles triangle --> <DOB=<DBO=30. <OBA=45-30=15 --> AOB is an isosceles triangle, so OA=OB. Also COB is an isosceles triangle, so CO=OB --> OA=CO=OB. So, AOC is an isosceles triangle --> <CAO=<OCA=45 (as <COA=90) --> x=45+30=75.

P.S. You can attach image files directly to the post.

Attachment:
Triangle complete.PNG

Hello sir,
Can you explain me by what properties you arrived at the solution?
By which property you drew CO such that it meets at the point where BO also meets to form such isosceles triangles?

Originally posted by ravi19012015 on 14 Feb 2017, 04:49.
Last edited by ravi19012015 on 14 Feb 2017, 11:32, edited 1 time in total.
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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14 Feb 2017, 11:04
Brunnel :

Could we have used a 45:45:90 for the triangle DOB by drawing DO perpendicular to AB. However, Doing this wont take us anywhere near the triangle ADC correct?

So. I guess my question is, Though I can understand by seeing values of sides and angles in triangles that I somehow have to use sides and angles combined to get my answer, and one of the most popular way of doing those in GMAT is using 45:45:90 or 30:60:90 triangles, but how do choose at one go which triangle should I use? What should be the thinking behind it? Which triangle to draw the perpendicular from among the trianlges ACD and ADB?
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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26 Jun 2017, 16:45
ODB is an isosceles triangle --> <DOB=<DBO=30. <OBA=45-30=15 --> AOB is an isosceles triangle, so OA=OB. Also COB is an isosceles triangle, so CO=OB --> OA=CO=OB. So, AOC is an isosceles triangle --> <CAO=<OCA=45 (as <COA=90) --> x=45+30=75.
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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09 Jul 2017, 17:04
navigator123 wrote:
enigma123 wrote:
In the figure, point D divides side BC of triangle ABC into segments BD and DC of lengths 1 and 2 units respectively. Given that ÐADC = 60º and ÐABD = 45º, what is the measure of angle x in degrees? (Note:
Figure is not drawn to scale.)

A. 55
B. 60
C. 70
D. 75
E. 90

Any idea guys what will be the correct answer please? Also any idea how can I cut and paste the pictures in my post? Is it possible?

Is it correct to tell that <CAD : <BAD = 2:1 because,
CD:DB = 2:1?

Even I have imagined in the same way and got 90 as the answer. I understood the explanation from the figures but my doubt is why can't <CAD = 30? when <DAB=15? If CD:DB=2:1, is it wrong to assume <CAD:<DAB is also 2:1??? please answer.

Thanks,
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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11 Jul 2017, 00:51
What is the source of this question?
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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11 Jul 2017, 00:57
rekhabishop wrote:
What is the source of this question?

The tag indicated that the source is Manhattan GMAT
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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21 Jul 2017, 00:54
Although I got the right answer, I wanted to understand simpler equation to derive the answer. Can someone help!
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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17 Aug 2017, 09:30
2
Attachment:

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. see attached pic

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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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06 Dec 2017, 20:10
Hi bunuel,

Sorry for bumping. A quick question.

How do we know that AD is not equal to 1. If it is equal to 1 then the ratio 2:1 when a perpendicular is drawn does not hold good right? Am i missing something?
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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06 Dec 2017, 20:40
gauthamvm wrote:
Hi bunuel,

Sorry for bumping. A quick question.

How do we know that AD is not equal to 1. If it is equal to 1 then the ratio 2:1 when a perpendicular is drawn does not hold good right? Am i missing something?

x=45+30=75.

Notes:
Sides with one blue segment crossing them are equal and sides with two blues segments crossing them are equal too.

CO is perpendicular to AD --> OD=1 (from 30-60-90 right triangle property as sides are in ratio $$1:\sqrt{3}:2$$) --> as OD=BD=1 then ODB is an isosceles triangle.
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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07 Dec 2017, 09:43
Hi Bunuel
is this a GMAT level question? seems too tough
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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14 Apr 2018, 11:57
Bunuel wrote:
enigma123 wrote:

In the figure, point D divides side BC of triangle ABC into segments BD and DC of lengths 1 and 2 units respectively. Given that ÐADC = 60º and ÐABD = 45º, what is the measure of angle x in degrees? (Note:
Figure is not drawn to scale.)

(A) 55
(B) 60
(C) 70
(D) 75
(E) 90

Any idea guys what will be the correct answer please? Also any idea how can I cut and paste the pictures in my post? Is it possible?

Complete solution for all the angles is in the image below:

x=45+30=75.

Notes:
Sides with one blue segment crossing them are equal and sides with two blues segments crossing them are equal too.

CO is perpendicular to AD --> OD=1 (from 30-60-90 right triangle property as sides are in ratio $$1:\sqrt{3}:2$$) --> as OD=BD=1 then ODB is an isosceles triangle.

<CDO and <BDO are supplementary to each other (supplementary angles are two angles that add up to 180°), so <BDO=120 --> <DAB=180-(120+45)=15.

As ODB is an isosceles triangle --> <DOB=<DBO=30. <OBA=45-30=15 --> AOB is an isosceles triangle, so OA=OB. Also COB is an isosceles triangle, so CO=OB --> OA=CO=OB. So, AOC is an isosceles triangle --> <CAO=<OCA=45 (as <COA=90) --> x=45+30=75.

P.S. You can attach image files directly to the post.

Attachment:
Triangle complete.PNG

Great Explanation!!
I didn't undertsand how is Triangle COB Isosceles??
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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15 Apr 2018, 10:19
vigrah wrote:
say angle CAB=y
since sum of angles in a triangle is 180
x+y+45=180
x+y=135 equation 1

line AD is dividing BC in 2:1 ratio
hence
X+2/3Y+60=180[u](can you explain this step please?

X+2/3y=120 equation 2
solving equation 1 &2 we get x=75
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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15 Apr 2018, 16:54
1
Hii
Is this GMAT level Question?
I am finding it tough to solve under time pressure in 2 minutes.
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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17 Apr 2018, 06:44
2
Top Contributor
enigma123 wrote:

In the figure, point D divides side BC of triangle ABC into segments BD and DC of lengths 1 and 2 units respectively. Given that ÐADC = 60º and ÐABD = 45º, what is the measure of angle x in degrees? (Note: Figure is not drawn to scale.)

A. 55
B. 60
C. 70
D. 75
E. 90

Attachment:
Triangle.png

Attachment:

First let's add a line from point C so that it is PERPENDICULAR to line AD
Also, note that, ∠EDB is 120° since it lies on a line with a 60° angle.

Since we already know 2 angles in ∆CED, we can see that the remaining angle, ∠ECD, is 30°, which means ∆CED is a 30-60-90 right triangle.
So, ∆CED is a 30-60-90 right triangle AND we know that the hypotenuse CD has length 2, we can conclude that side ED has length 1.

Now draw a line from E to B.
Since ED = EB (both have length 1), we can see that ∆EDB is an ISOSCELES TRIANGLE, which means the two remaining angles are each 30°

Next, ∠EAB is 150° since it shares a line with a 30° angle
Also, since we were originally told that ∠CBA is 45°, we can conclude that ∠ABE is 15°

Now focus on ∆CBE
Notice that this triangle is an ISOSCELES triangle, because ∠ECB = ∠EBC = 30°
This means that side EC = side EB

Now focus on ∆EAB
Since we already know two of the angles in this triangle (150° and 15°), we can conclude that = ∠EAB = 15°

Stay focused on ∆EAB
This triangle is an ISOSCELES triangle, because ∠EAB = ∠EBA = 15°
This means that side AE = side EB

Now focus on ∆ACE
Since CE = EA, this is an ISOSCELES triangle
Since one angle = 90°, the other two angles are each 45°

At this point, we can see that x = 45° + 30° = 75°

Cheers,
Brent
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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29 May 2018, 02:49
1
vigrah wrote:
say angle CAB=y
since sum of angles in a triangle is 180
x+y+45=180
x+y=135 equation 1

line AD is dividing BC in 2:1 ratio
hence
X+2/3Y+60=180
X+2/3y=120 equation 2
solving equation 1 &2 we get x=75

This answer is wrong. Solving this system does not yield x=75, but x=90, and y=45.
This answer should not have so many kudos, it is very misleading.
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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20 Jun 2018, 09:20
danzig wrote:
Bunuel,

How can we know that we had to draw a perpendicular line from vertex C? , Why not from vertex D?, for example?

Thanks!

Posted from my mobile device
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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31 Jul 2018, 07:54
vigrah wrote:
say angle CAB=y
since sum of angles in a triangle is 180
x+y+45=180
x+y=135 equation 1

line AD is dividing BC in 2:1 ratio
hence
X+2/3Y+60=180
X+2/3y=120 equation 2
solving equation 1 &2 we get x=75

Hi, Can you please explain how you arrived at X + 2/3Y +60 = 180????
Re: In the figure, point D divides side BC of triangle ABC into segments   [#permalink] 31 Jul 2018, 07:54

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