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

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

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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

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?

[Reveal] Spoiler:
Attachment:
Triangle.png
Triangle.png [ 9.63 KiB | Viewed 57529 times ]
[Reveal] Spoiler: OA

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Last edited by Bunuel on 15 Sep 2016, 04:29, edited 2 times in total.
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In the figure, point D divides side BC of triangle ABC into [#permalink]

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enigma123 wrote:
Image

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:

Image

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.

Answer: D.

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

[Reveal] Spoiler:
Attachment:
Triangle complete.PNG
Triangle complete.PNG [ 36.32 KiB | Viewed 58076 times ]

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

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Could you help to understand this problem please?
I've come out with E

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

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

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New post 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.

Answer: D.

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 [#permalink]

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New post 18 Sep 2012, 01:13
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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.

Answer: D.

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.


Not sure I understand your question. CO is perpendicular to AD because we made it so.
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Collection of Questions:
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Re: In the figure, point D divides side BC of triangle ABC into [#permalink]

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New post 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 [#permalink]

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New post 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.

Answer: D.

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.

Not sure I understand your question. CO is perpendicular to AD because we made it so.


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 [#permalink]

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New post 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.

Answer: D.

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
<CAD = 180 -60 -x
<CAD = 120 -x

For the other triangle we know
<DAC = 15
<ADB = 120
<DBA =45

Exterior angle is sum of opp interior angle and not adjacent to it
So <CAD = <ADB + <DBA
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 [#permalink]

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New post 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.

Answer: D.

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
<CAD = 180 -60 -x
<CAD = 120 -x

For the other triangle we know
<DAC = 15
<ADB = 120
<DBA =45

Exterior angle is sum of opp interior angle and not adjacent to it
So <CAD = <ADB + <DBA
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
Exterior.png [ 44.25 KiB | Viewed 53506 times ]
Hope it's clear.
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Re: In the figure, point D divides side BC of triangle ABC into [#permalink]

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

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New post 20 Jun 2013, 04:04
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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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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

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New post 24 Jun 2013, 11:41
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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
answer is D

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

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New post 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.

Answer: D.
.


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 [#permalink]

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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.

Answer: D.
.


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 [#permalink]

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New post 11 Aug 2013, 11:15
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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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.

Answer: D.
.


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
TRIANGLE.png [ 39.43 KiB | Viewed 51770 times ]


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

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New post 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.

Answer: D.
.


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 [#permalink]

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New post 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 [#permalink]

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New post 19 May 2014, 07:31
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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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: In the figure, point D divides side BC of triangle ABC into   [#permalink] 19 May 2014, 07:31

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