Geometry determining solution using law of sines

This is an SSA problem, so we have to be mindful of the ambiguous case. I would begin by writing:

[MATH]\frac{\sin(50^{\circ})}{34}=\frac{\sin(B)}{40}\implies B=\arcsin\left(\frac{20}{17}\sin(50^{\circ})\right)\approx64.3^{\circ}[/MATH]
Thus, we must also consider:

[MATH]B=180^{\circ}-\arcsin\left(\frac{20}{17}\sin(50^{\circ})\right)\approx115.7^{\circ}[/MATH]
Given this, can you determine the two possible values for \(\angle C\) and side \(c\)?
 
MarkFL said:
This is an SSA problem, so we have to be mindful of the ambiguous case. I would begin by writing:

[MATH]\frac{\sin(50^{\circ})}{34}=\frac{\sin(B)}{40}\implies B=\arcsin\left(\frac{20}{17}\sin(50^{\circ})\right)\approx64.3^{\circ}[/MATH]
Thus, we must also consider:

[MATH]B=180^{\circ}-\arcsin\left(\frac{20}{17}\sin(50^{\circ})\right)\approx115.7^{\circ}[/MATH]
Given this, can you determine the two possible values for \(\angle C\) and side \(c\)?
Click to expand...
Are my calculations correct?
 
rachelmaddie said:
I don’t understand what you did to get 20/17?
Click to expand...

Let's go back to:

[MATH]\frac{\sin(50^{\circ})}{34}=\frac{\sin(B)}{40}[/MATH]
We want to solve for \(B\), so let's multiply by 40:

[MATH]\frac{40\sin(50^{\circ})}{34}=\sin(B)[/MATH]
Arrange as:

[MATH]\sin(B)=\frac{40}{34}\sin(50^{\circ})[/MATH]
Reduce the fraction on the RHS:

[MATH]\sin(B)=\frac{20}{17}\sin(50^{\circ})[/MATH]
Thus:

[MATH]B=\arcsin\left(\frac{20}{17}\sin(50^{\circ})\right)[/MATH]
 
MarkFL said:
Let's go back to:

[MATH]\frac{\sin(50^{\circ})}{34}=\frac{\sin(B)}{40}[/MATH]
We want to solve for \(B\), so let's multiply by 40:

[MATH]\frac{40\sin(50^{\circ})}{34}=\sin(B)[/MATH]
Arrange as:

[MATH]\sin(B)=\frac{40}{34}\sin(50^{\circ})[/MATH]
Reduce the fraction on the RHS:

[MATH]\sin(B)=\frac{20}{17}\sin(50^{\circ})[/MATH]
Thus:

[MATH]B=\arcsin\left(\frac{20}{17}\sin(50^{\circ})\right)[/MATH]
Click to expand...
So how do I determine the solution?
 
rachelmaddie said:
So how do I determine the solution?
Click to expand...

I gave you two possible values for \(\angle B\):

i) [MATH]B=\arcsin\left(\frac{20}{17}\sin(50^{\circ})\right)[/MATH]
ii) [MATH]B=180^{\circ}-\arcsin\left(\frac{20}{17}\sin(50^{\circ})\right)[/MATH]
For each of these cases, what is the value of \(\angle C\)? And then, using the Law of sines, what are the corresponding values for side \(c\)?
 
MarkFL said:
I gave you two possible values for \(\angle B\):

i) [MATH]B=\arcsin\left(\frac{20}{17}\sin(50^{\circ})\right)[/MATH]
ii) [MATH]B=180^{\circ}-\arcsin\left(\frac{20}{17}\sin(50^{\circ})\right)[/MATH]
For each of these cases, what is the value of \(\angle C\)? And then, using the Law of sines, what are the corresponding values for side \(c\)?
Click to expand...
Okay, I’ve never worked with reducing fractions in this case?
 
MarkFL said:
It's not necessary to reduce fractions to find the angles, but it's just good form to do so.
Click to expand...
Well, so far I have m<C = 180 - (50 + 64.3) = 114.3
m<C = 180 - (50 + 115.7) = 14.3
Sin 114.3/C = sin50/34
34sin114.3/50 = 1.4
 
For the first case I listed, we could state:

[MATH]C=180^{\circ}-\left(50^{\circ}+\arcsin\left(\frac{20}{17}\sin(50^{\circ})\right)\right)=130^{\circ}-\arcsin\left(\frac{20}{17}\sin(50^{\circ})\right)\approx65.7^{\circ}[/MATH]
And then using the Law of Sines:

[MATH]c=\frac{34\sin(C)}{\sin(50^{\circ})}\approx?[/MATH]
For the second case:

[MATH]C=180^{\circ}-\left(50^{\circ}+180^{\circ}-\arcsin\left(\frac{20}{17}\sin(50^{\circ})\right)\right)=\arcsin\left(\frac{20}{17}\sin(50^{\circ})\right)-50^{\circ}\approx14.3^{\circ}[/MATH]
And then using the Law of Sines:

[MATH]c=\frac{34\sin(C)}{\sin(50^{\circ})}\approx?[/MATH]
 
As a general rule, I would advise against using the values rounded to one decimal place in intermediary calculations. :)
 
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