/sci/ - Science & Math » Thread #11018619

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Anonymous Mon 30 Sep 01:08:55 2019 No.11018619 View Reply Original Report

Quoted By: >>11019207 >>11019285 >>11019292

In the triangle ABC, the median CX is perpendicular to the bisectrix AY.
CM:AL is known.
How do you find the angle BAC using vectors?

Anonymous

Anonymous Mon 30 Sep 2019 01:11:08 No.11018631 Report

Quoted By:

I meant CX:AY is known

Anonymous

Anonymous Mon 30 Sep 2019 03:03:03 No.11018925 Report

Quoted By:

bump

Anonymous

Anonymous Mon 30 Sep 2019 03:14:01 No.11018941 Report

Quoted By:

nobody is going to do your homework for you if you cant even state the problem correctly you lazy mouth breathing retard.

Anonymous

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Anonymous Mon 30 Sep 2019 06:15:18 No.11019207 Report

Quoted By:

>>11018619
Is this your problem?

Anonymous

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Anonymous Mon 30 Sep 2019 07:22:55 No.11019285 Report

Quoted By:

>>11018619
First, notice that triangles CAF and XAF are congruent. Then apply the angle bisector theorem to triangle ABC to conclude YC/YB =1/2. Then apply Menalaus Theorem on AXBYCFX to get YF/FA = 1/3. Then you can solve for n in terms of m from the fact that CFA and CFY are right triangles sharing the same height. After that just apply the law of cosine to solve for the angle

Anonymous

Anonymous Mon 30 Sep 2019 07:30:31 No.11019292 Report

Quoted By:

>>11018619
I guess you could also use Stewart's theorem from the two cevians you have and set up two equations, and from the ratio CX/AY you could solve for n in terms of m and that would give you the same answer.

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