Did the problem state AD =6 or AC=6 ?
I've been introduced to the Law of Tangents and I have some doubts about it.
In , let D be a point in BC such that AD bisects angle A. Given that AD=6, BD=4, and DC=3, find AB.
Is there any other possibility?
Is the Law of Tangents really needed here? If so, how do we use it?
No, that's all!
The question is here Law of Tangents - AoPSWiki
It was a question of Mu Alpha Theta. They have really great questions.
Unfortunately, I can't find the solved test from 1991.
It involves the Law of Tangents, nevertheless.
Wilmer shows the use of the angle bisector theorem which is what I thought the problem was supposed to be.I looked at your reference which shows it as unedited.The problem is trig not geometry,My own opinion is that the law of tangents is another way to solve a triangle given two sides and included angle (rather than cosine law)
I had to edit the post several times, because it was actually wrong.
The right is CB/AC=AC/CD.
If we consider that the angles ADB and angle ADC are supplementary; their sines will be equal.
And since we have AB/BD=sin angle ADB/sin angle DAB and AC/DC=sin angle ADC/sin angle DAC.
Or something along this lines.
Thanks bjhopper and Wilmer!
I got it now, it was a minor mistake here.
That is not how I usually think about a problem. I need to have some base in knowledge before I can TRY anything.
Therefore, I never ASSUME something is right without proof.
OKkkkkk....if you're "collecting" interesting ones:
triangle ABC with sides BC = 24, AC = 40 and AB = 56
D on BC, E on AC, F on AB such that AD, BE and CF are angle bisectors
BC: BD = 14, CD = 10
AC: AE = 28, CE = 12
AB: AF = 35, BF = 21
In other words, ALL integers.
Have fun(?): try case where triangle ABC = 70-105-140.
Sorry, I had to resuscitate this thread. Since I still don't get it, really sorry about that.
The question is, how does comes into play?
I can only find the bisector theorem in this terms...
I don't know what the the length of the angle bisector equals so it I really don't know how to use it.
It is present in Stewart's Theorem, but to use it we would need another leg.
Can someone do it step by step? That would be really appreciated. Thanks.
If I understand your new post you want to know how to find the lengths of AB and AC using Stewards Theorem. I am not familar with it but I have a plan for you.
Draw a vertical line length 6 Mark upper end A and lower D.At point D swing two arcs lengths of 3 and 4.A straight line thru D defines two points B and C.Note that you can draw many triangles.If you fix the length of either AB or AC the other line can be calculated using the cosine law