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

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Manager
Joined: 08 Jul 2018
Posts: 66
Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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05 Oct 2018, 02:12
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

How you took the value of angle CAD as 2/3y?
Intern
Joined: 13 Jul 2018
Posts: 10
Re: In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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17 Oct 2018, 05:36
In this why can't we take angle ADB = 120 degrees which can provide angle DAB = 15 degrees and since BD = 1/2 CD so DAC = 30 degrees.

What's wrong in this approach?
Manager
Joined: 10 Aug 2009
Posts: 63
In the figure, point D divides side BC of triangle ABC into segments  [#permalink]

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20 Dec 2018, 12:30
arpitalewe wrote:
In this why can't we take angle ADB = 120 degrees which can provide angle DAB = 15 degrees and since BD = 1/2 CD so DAC = 30 degrees.

What's wrong in this approach?

This is wrong. The property you are referring works for angles and sides within a triangle but here you are comparing two different triangles. However you can use the angle bisector theorem here, according to which:

$$CD/DB = sin(angle DAC)/sin (angle DAB)$$

Now you need to know the value of sin15 and find the double of it and then find the angle whose value is represented by that sin value. Here is the link from wiki for angle bisector theorem:
https://en.wikipedia.org/wiki/Angle_bisector_theorem
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In the figure, point D divides side BC of triangle ABC into segments &nbs [#permalink] 20 Dec 2018, 12:30

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