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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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Question Stats:

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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 96039 times ]

Attachment:

task%236.png [ 50.38 KiB | Viewed 11404 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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Joined: 02 Sep 2009
Posts: 48037
Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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01 Feb 2012, 17:57
60
56
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 97023 times ]

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

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24 Jun 2013, 11:41
23
7
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
##### General Discussion
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Joined: 23 Aug 2012
Posts: 12
Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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18 Sep 2012, 00:07
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:
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.

Thanks for the explanation Bunuel. However, I just want to ask for a bit of a clarification. I understand how we create the 30-60-90 triangle to find the solution, but we only do this because the problem clues us in to try this out, not because it has to be true. For example, what if x was less than 30? All we have is B=45 and A=15+some angle. The problem says figure is not drawn to scale, so why should we assume that CO is perpendicular to AD? We don't actually know that any of our assumptions hold until the very end when we see that all of the sides and angles do, in fact, work out, right?

I hope my confusion makes some sense. I guess I tend to have a problem with making such assumptions, especially when the figure is not drawn to scale. I always am afraid that I'm going to assume something that isn't necessarily true.
Math Expert
Joined: 02 Sep 2009
Posts: 48037
Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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18 Sep 2012, 01:13
1
dandarth1 wrote:
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:
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.

Thanks for the explanation Bunuel. However, I just want to ask for a bit of a clarification. I understand how we create the 30-60-90 triangle to find the solution, but we only do this because the problem clues us in to try this out, not because it has to be true. For example, what if x was less than 30? All we have is B=45 and A=15+some angle. The problem says figure is not drawn to scale, so why should we assume that CO is perpendicular to AD? We don't actually know that any of our assumptions hold until the very end when we see that all of the sides and angles do, in fact, work out, right?

I hope my confusion makes some sense. I guess I tend to have a problem with making such assumptions, especially when the figure is not drawn to scale. I always am afraid that I'm going to assume something that isn't necessarily true.

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

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18 Sep 2012, 11:44
I guess what I'm saying is, what if CA was perpendicular to AD? Then making a new point CO that is perpendicular to AD couldn't be possible, unless O was equal to A. Therefore, aren't we making an unwarranted assumption that CA is not perpendicular to AD?
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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18 Sep 2012, 11:46
Bunuel wrote:
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:
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.

Thanks for the explanation Bunuel. However, I just want to ask for a bit of a clarification. I understand how we create the 30-60-90 triangle to find the solution, but we only do this because the problem clues us in to try this out, not because it has to be true. For example, what if x was less than 30? All we have is B=45 and A=15+some angle. The problem says figure is not drawn to scale, so why should we assume that CO is perpendicular to AD? We don't actually know that any of our assumptions hold until the very end when we see that all of the sides and angles do, in fact, work out, right?

I hope my confusion makes some sense. I guess I tend to have a problem with making such assumptions, especially when the figure is not drawn to scale. I always am afraid that I'm going to assume something that isn't necessarily true.

I guess what I'm saying is, what if CA was perpendicular to AD? Then making a new point CO that is perpendicular to AD couldn't be possible, unless O was equal to A. Therefore, aren't we making an unwarranted assumption that CA is not perpendicular to AD?
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Joined: 22 Dec 2011
Posts: 267
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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Joined: 02 Sep 2009
Posts: 48037
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 84110 times ]
Hope it's clear.
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Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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19 Jun 2013, 04:31
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?
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Joined: 02 Sep 2009
Posts: 48037
Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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20 Jun 2013, 04:04
1
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?

Nope, that's not correct. Check the images provided above.
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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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Posts: 48037
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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11 Aug 2013, 11:15
1
Bunuel,

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

Thanks!
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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
9
5
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 82148 times ]

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Asif vai.....

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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, 12:39
Asifpirlo wrote:
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:

Hmmmmm hope this can clarify the fact..........................
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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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Posts: 48037
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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19 May 2014, 08:02
enigma123 wrote:
Attachment:
Triangle.png
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?

Another approach can be used without using constructions:

Angle DAB = 15
CAD = 180 - 60 - x = 120 -x

Now using sine rules first in triangle ACD and then in triangle ADB

sin x/AD = sin(120-x)/2t ----- 1

sin 45/AD = sin 15/t -----------2

Dividing 1 by 2

sin x/sin 45 = sin(120 -x)/ 2sin 15

Now using options, we need to check:

Now if you put 75 in this, you will see that the equality holds true.

This is a tricky solution though, but if you use a little bit of intuition, you may arrive at a right answer.
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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 2014, 01:05
1
Since DB and CD are in the ratio 1:2 i.e. db= 1/3 of CB & CD= 2/3 of CB. the angle will also be in proportionate value.
angle DAC=1/3 angle CAB.

Now, angle ADB= 180-60 = 120 deg.
=> angle DAB = 15 deg.

So, angle CAB= 45 deg. => angle ACB = 90 deg.
Re: In the figure, point D divides side BC of triangle ABC into segments &nbs [#permalink] 15 Nov 2014, 01:05

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