Law of sines and triangle solutions

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Homework Help Overview

The discussion revolves around a triangle ABC where angle A is given as 60 degrees, the length of side BC is the square root of 3, and side AC measures 1/5. Participants are exploring how many solutions exist for the measure of angle B using the Law of Sines.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the application of the Law of Sines and the implications of their calculations, questioning the validity of their arithmetic and the resulting sine values. There is also exploration of the ambiguous case in triangle solutions and the potential for multiple angles fitting the criteria.

Discussion Status

Some participants are attempting to clarify their calculations and reasoning, while others suggest drawing diagrams to visualize the problem better. There is an ongoing exploration of how to determine the number of solutions based on the calculated angles and the properties of triangles.

Contextual Notes

There is uncertainty regarding the assignment of the base of the triangle and the implications of the undefined sine values. Participants are also considering the constraints of the arcsin function and the potential for multiple solutions in the context of the ambiguous case.

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Homework Statement


Given triangle ABC with the measure of angel A = 60 degrees, the length of BC = sq rt of 3, and the length of AC = 1/5. How many soulutions are there for the measure of angle B?

The Attempt at a Solution


1. sin(60) / sq rt 3 = sinB / (1/5)

2. sq rt 3 / 2 * 1 / sq rt 3 * 5

3. 5/2 = sin(B)
= Undefined because domain is [-1,1]

It would seem I did something wrong but I keep getting the same thing, any assistance would be great.
 
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2. sq rt 3 / 2 * 1 / sq rt 3 * 5


Should be:


2. sq rt 3 / 2 * 1 / sq rt 3 * (1/5)


You have just made an arithmetic mistake.
 
So if angle B is 5.74 degrees how would you go about finding how many solutions there are? Originally I tried seeing how many times 5.8 went into 90 but I decided that couldn't be right so I started to subtract it from 180 degrees and see if it fit in the domain of arcsin. Which also seems to have provided the wrong answer :(. If the answer were undefined would the number of solutions be zero?
 
This is the ambiguous case for the Law of Sines -- A S S. ;)

I think you're going about this the wrong way -- I'm not sure you're supposed to find angle B. First, draw a diagram (hint: make AB the base of the triangle). Drop a perpendicular from C to side AB -- this is the height of the triangle (h). How would you go about finding h?01
 
Well assuming that is possible and setting AB as base ( I am not sure how that is assigned the base ) I get..

1. Angle A = 60 Angle B = 5.74 therefore Angle C = 114.26
2. Draw perpendicular and if base is DE now in triangle DEF then Angle F is 57.13
3. 180 - (57.13 + 90 = 147.13) = 32.87 Therefore Angle E is 32.87
4. sin(32.87)= X / sq rt 3 Thus X(h) = 0.94
5.Then from that use a2+b2=c2 to get 1/2 base which ended up being 0.92 so base = 1.84
6.AB = 1.84

Well the problem is I do not know how to find the number of solutions from this either ;/.

The answer is either 0,1,2,3, or can not be determined as I as I already missed it with none of the above. =/ It would seem about 60% of time the answer is two but its not very helpful for when its not lol
 
I'm not sure what you did ^ up there. Can you draw a diagram and post it here?

I don't think you got the height of the triangle (h as I defined it). Hint: the way to find h doesn't involve the Law of Sines at all.


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