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

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

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19 May 2014, 07:02
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 [#permalink]

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15 Nov 2014, 00:05
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.
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Re: In the figure, point D divides side BC of triangle ABC into [#permalink]

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04 Feb 2015, 13:54
So easy if you look at it the right way...
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Re: In the figure, point D divides side BC of triangle ABC into [#permalink]

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06 Mar 2015, 17:47
enigma123 wrote:
Attachment:
Triangle.png

Guys,
Is there any relation between the ratio of the sides to the angle containing them ?

i. e from the diagram ,$$\frac{CD}{DB}$$ = ratio of $$\angle CAD$$ and $$\angle DAB$$ ?
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Re: In the figure, point D divides side BC of triangle ABC into [#permalink]

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10 May 2015, 12:24
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

According to your calculation x=90 not 75
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Re: In the figure, point D divides side BC of triangle ABC into [#permalink]

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16 May 2015, 06:44
@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 [#permalink]

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18 May 2015, 10:45
1
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Expert's post
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
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Re: In the figure, point D divides side BC of triangle ABC into [#permalink]

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18 May 2015, 10:50
3
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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 2747 times ]

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

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28 May 2015, 11:37
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

That looks like a very efficient way to solve this. Could you explain the red formatted part? How do you set up this equation?
Thank you
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Re: In the figure, point D divides side BC of triangle ABC into [#permalink]

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28 May 2015, 11:39
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 [#permalink]

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28 May 2015, 20:13
1
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Expert's post
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 [#permalink]

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28 May 2015, 20:34
<DAB can be calculated to be 15 degree
The side AD divides <CAB in the same ratio as it divides the side CB i.e. 2:1
Thus - <CAD = 30 degree and x = 90 degree
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Re: In the figure, point D divides side BC of triangle ABC into [#permalink]

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12 Jul 2015, 06:11
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 [#permalink]

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12 Jul 2015, 06:55
1
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This post was
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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 [#permalink]

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27 Oct 2015, 18:39
bunnuel's explanation is brilliant.
I did a mistake, and drew a perpendicular to CD, and concluded that angle A must be 60...
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Re: In the figure, point D divides side BC of triangle ABC into [#permalink]

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24 Nov 2015, 20:04
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

x+y = 135---1
x+2/3y = 120----2

subtracting 2 from 1

1/3 y = 15

y =45.

Substituting y =45 in eq 1
x+45 = 135
x = 90.

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

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07 Mar 2016, 03:35
Let me provide you an easy approach for this question:

Using side & angle ratio to determine x

assume angle DAC = y

then, angle (BAD/BAC) = side (BD/BC)

=> 15/y = 1/(1+2)
=> 15/y = 1/3
=> 15*3 = y
=>45

Now, as we know sum of all angles of a triangle is 180
Therefore, in triangle ADC, Angle (DAC+ACD+CDA) = 180
=> 60+45+X = 180
=> X = 75
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Re: In the figure, point D divides side BC of triangle ABC into [#permalink]

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15 Mar 2016, 10:13
Since the AD divides the BC in ration 2:1 , it means if we extend the traingle and make it eqilateral triangle. The medians of equilateral triangles divides each other in ration 2:1 . Thus easily all sides are 60 degrees and the x = 75. ..Simple. Please give Kudos .
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Re: In the figure, point D divides side BC of triangle ABC into [#permalink]

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11 Aug 2016, 13:11
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.
Attachments

example2.png [ 13.01 KiB | Viewed 1441 times ]

Last edited by OmnerLV on 12 Aug 2016, 19:26, edited 1 time in total.
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Re: In the figure, point D divides side BC of triangle ABC into [#permalink]

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11 Aug 2016, 15:19
here are my way to solve it:
Attachments

mi-proceso.png [ 40.48 KiB | Viewed 1383 times ]

Re: In the figure, point D divides side BC of triangle ABC into   [#permalink] 11 Aug 2016, 15:19

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