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

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

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Updated on: 06 Apr 2018, 04:44
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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 [ 9.63 KiB | Viewed 122797 times ]

Attachment:

task%236.png [ 50.38 KiB | Viewed 31885 times ]

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Originally posted by enigma123 on 01 Feb 2012, 16:32.
Last edited by Bunuel on 06 Apr 2018, 04:44, edited 3 times in total.
Attached the image.
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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01 Feb 2012, 17:57
68
67
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 [ 36.32 KiB | Viewed 120581 times ]

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

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13 Aug 2013, 09:57
11
8
Transcendentalist wrote:
Bunuel wrote:
--> 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.

.

I lost you from here. How do you know that angle cab is 45+15?

step by step using this diagram can help to understand:
Attachments

TRIANGLE.png [ 39.43 KiB | Viewed 102479 times ]

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

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03 Nov 2012, 02:10
Bunuel wrote:
Complete solution for all the angles is in the image below:
Attachment:
Triangle complete.PNG
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.

Bunuel - Can we not use exterior angle property

for triangle ACD
Angle at C = x
<CDA = 60
Since sum of all the angle of triangle is 180 then we

<CAD + <CDA + x = 180

For the other triangle we know
<DAC = 15
<DBA =45

Exterior angle is sum of opp interior angle and not adjacent to it
120 -x = 120 + 45
x = 45.

Since this is not the ans could you please let me know what I'm doing wrong here?

Many thanks.

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

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03 Nov 2012, 02:23
Jp27 wrote:
Bunuel wrote:
Complete solution for all the angles is in the image below:
Attachment:
The attachment Triangle complete.PNG is no longer available
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.

Bunuel - Can we not use exterior angle property

for triangle ACD
Angle at C = x
<CDA = 60
Since sum of all the angle of triangle is 180 then we

<CAD + <CDA + x = 180

For the other triangle we know
<DAC = 15
<DBA =45

Exterior angle is sum of opp interior angle and not adjacent to it
120 -x = 120 + 45
x = 45.

Since this is not the ans could you please let me know what I'm doing wrong here?

Many thanks.

Cheers

<CAD is not exterior angle of <DAB. Exterior angle for <DAB is given below in green:
Attachment:

Exterior.png [ 44.25 KiB | Viewed 104441 times ]
Hope it's clear.
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In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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24 Jun 2013, 11:41
54
18
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

Edit: THIS SOLUTION IS NOT CORRECT. IGNORE IT
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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25 Jun 2013, 10:10
Bunuel wrote:
--> 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.

.

I lost you from here. How do you know that angle cab is 45+15?
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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25 Jun 2013, 10:53
1
Transcendentalist wrote:
Bunuel wrote:
--> 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.

.

I lost you from here. How do you know that angle cab is 45+15?

<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.

<CAB = <DAB + <CAO = 15 + 45.
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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19 May 2014, 07:05
hi Bunnel,

How can we be sure that AO is equals to CO?
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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19 May 2014, 07:31
1
pretzel wrote:
hi Bunnel,

How can we be sure that AO is equals to CO?

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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16 May 2015, 07:44
@Bunnel

Can you please explain how this is derived?

"As 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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18 May 2015, 11:50
3
ashtitude wrote:
@Bunnel

Can you please explain how this is derived?

"As ODB is an isosceles triangle"

May be the attached image can help explain how.
Attachments

IMG-20150514-WA0014.jpg [ 44.46 KiB | Viewed 7518 times ]

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

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28 May 2015, 12:39
EgmatQuantExpert wrote:
ashtitude wrote:
@Bunnel

Can you please explain how this is derived?

"As ODB is an isosceles triangle"

Hi ashtitude,

Using image drawn by Bunuel for explanation here.

You can observe in the figure that triangle OCD is a 30-60-90 triangle, hence it sides will be in the ratio 1:√3:2. Since CD = 2, we will have OD = 1. So in triangle OBD we have OD = BD = 1 which makes ODB an isosceles triangle.

Hope this helps

Regards
Harsh

Nice explanation. What confuses me more is, how do we know that AOB is isosceles?
Thank you
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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28 May 2015, 21:13
1
reto wrote:
EgmatQuantExpert wrote:
ashtitude wrote:
@Bunnel

Can you please explain how this is derived?

"As ODB is an isosceles triangle"

Hi ashtitude,

Using image drawn by Bunuel for explanation here.

You can observe in the figure that triangle OCD is a 30-60-90 triangle, hence it sides will be in the ratio 1:√3:2. Since CD = 2, we will have OD = 1. So in triangle OBD we have OD = BD = 1 which makes ODB an isosceles triangle.

Hope this helps

Regards
Harsh

Nice explanation. What confuses me more is, how do we know that AOB is isosceles?
Thank you

Hi reto,

In triangle ADB, ∠ADB = 120 ( as∠ADB = 180 - ∠CDO = 180 - 60 =120). We now know two angles of the triangle ADB. So the third angle ∠DAB = 180 -120 -45 = 15.

Similarly in triangle ODB since ∠DBO =30 (as ODB is an isosceles triangle), we have ∠OBA = 45 - 30 = 15.

So we see that ∠OBA = ∠DAB = 15. These are two angles of triangle AOB. Hence AOB is an isosceles triangle with sides OA = OB

Hope it's clear

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

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12 Jul 2015, 07:11
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 looks much more simple!

Just so I am clear because line AD bisects line CB in a two to 1 ratio we can calculate the breakdown of angles at vertex A?

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

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12 Jul 2015, 07:55
1
DropBear wrote:
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 looks much more simple!

Just so I am clear because line AD bisects line CB in a two to 1 ratio we can calculate the breakdown of angles at vertex A?

This is a very dangerous inference. The only formula that links sides to angles in by sine law from trigonometry that says:

In a triangle ABC with sides a,b,c opposite $$\angle (A), \angle (B) and \angle (C)$$ respectively, then

a/sin(A) = b/sin(B) = c/sin(C). You do not have to remember this 'sine law' for GMAT.

It was a lucky coincidence that the user was able to arrive at the correct answer by assuming that the angles themselves got divided into 1:2 ratio if the sides were divided in the ratio 1:2.

The best way would be to follow what Bunuel did in his approach and for obtaining $$\angle (BAD) and \angle (CAD)$$, after you find OD = DB = 1 ---> $$\angle {DBO} = \angle {DOB}$$ = 30 degrees.

Also, as we know $$\angle{ABD}$$ = 45 degrees ---->$$\angle {ABO}$$ = 15 degrees. Additionally, $$\angle{ADB} = \angle {ADB} + \angle {DBO}$$ ... (by external angle theorem of triangles).

Once you do this, you will see $$\angle{BAO}$$ = 15 degrees. Then , you proceed to finding other angles as stated by Bunuel.
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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: 12 Aug 2016, 20:26
We are solving it with the misconception that if we make a perpendicular from the point C to the line AD will perfectly connect with the point O(M). In this case fortunately matched, but if you try to use the same logic just modifying the value of line CD to 3, this method won't work. We need to find a real answer.

-UPDATE-

before start working with the original triangle, we need to understand the equilateral triangle properties. I'll explain later.
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example2.png [ 13.01 KiB | Viewed 6093 times ]

Originally posted by OmnerLV on 11 Aug 2016, 14:11.
Last edited by OmnerLV on 12 Aug 2016, 20:26, 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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11 Aug 2016, 16:19
here are my way to solve it:
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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12 Aug 2016, 21:28
1
A GMAT way to answer the problem, the best way I can explain it. (click on the image and then in the "magnifying glass" at the top left to enlarge to full size)
Don't forget +1KUDOS !!!
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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23 Aug 2016, 22:54
vigrah wrote:
say angle CAB=y
since sum of angles in a triangleis 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 DOES NOT give us the correct solution. There is a calculation mistake here. Using the calculation mentioned, we get x=90 which is incorrect. This basically implies that the statement
"line AD is dividing BC in 2:1 ratio - > therefore corresponding angles also get divided in the same proportion" is invalid. The only way to go forward is to follow the approach that Bunuel mentioned.

Hope this helps clear the confusion around this.

Thanks.
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Re: In the figure, point D divides side BC of triangle ABC into segments   [#permalink] 23 Aug 2016, 22:54

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