Law of sines and triangle solutions

In summary: 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 = 114.262. Draw perpendicular and if base is DE now in triangle DEF then Angle F is 57.133. 180 - (57.13 + 90 = 147.13) = 32.87 Therefore Angle E is 32.874. sin(32.87)= X / sq rt 3 Thus X(h) = 0.945.Then from that use a2+b2=c2 to get 1/2 base which ended up being 0.92 so base =
  • #1
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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

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.
 
  • #3
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?
 
  • #4
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
 
  • #5
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
 
  • #6
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
 

1. What is the Law of Sines and how is it used to solve triangle problems?

The Law of Sines is a trigonometric rule that relates the lengths of a triangle's sides to the sine of its angles. It states that the ratio of the length of a side to the sine of its opposite angle is constant for all sides and angles in a given triangle. This law is used to solve triangle problems by finding missing side lengths or angles when given enough information about the other sides and angles.

2. When can the Law of Sines be used to solve a triangle?

The Law of Sines can be used to solve a triangle when at least one angle and its opposite side are known, or when two angles and one side are known. This is because the law requires at least one pair of a side and its opposite angle to be known in order to find the ratio between them.

3. How do you use the Law of Sines to find a missing angle or side length in a triangle?

To use the Law of Sines to find a missing angle or side length, you would first identify which sides and angles are known. Then, you would set up a proportion using the known values and the Law of Sines formula, which is sin(A)/a = sin(B)/b = sin(C)/c. Finally, you can solve for the missing value by cross-multiplying and taking the inverse sine of the resulting ratio.

4. Can the Law of Sines be used for any type of triangle?

Yes, the Law of Sines can be used for any type of triangle, including acute, right, and obtuse triangles. This is because the law is based on the relationship between the lengths of sides and the sine of angles, which is constant for all types of triangles.

5. Are there any limitations to using the Law of Sines to solve triangles?

There are a few limitations to using the Law of Sines. First, the law can only be used when there is enough information given about the triangle, such as at least one angle and its opposite side. Additionally, the law may not be accurate for very small or very large angles, as the sine function becomes less precise at extreme values. Lastly, the law is not applicable for solving right triangles, as the sine of a right angle is undefined.

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