Law of sines and triangle solutions

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SUMMARY

The discussion centers on solving triangle ABC using the Law of Sines, given angle A = 60 degrees, side BC = √3, and side AC = 1/5. The user initially misapplies the Law of Sines, leading to an undefined value for sin(B). The correct approach involves drawing a diagram, dropping a perpendicular from C to side AB to find the height (h), and then calculating the angles and sides accordingly. The ambiguous case of the Law of Sines is highlighted, indicating that the number of solutions for angle B can be 0, 1, or 2, depending on the configuration of the triangle.

PREREQUISITES
  • Understanding of the Law of Sines
  • Basic trigonometric functions and their domains
  • Ability to draw and interpret geometric diagrams
  • Familiarity with triangle properties and angle relationships
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  • Study the ambiguous case of the Law of Sines in detail
  • Learn how to derive triangle heights using trigonometric functions
  • Practice solving triangles with given angles and sides using the Law of Sines
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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.


01
 

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