Geometry determining solution using law of sines

[rachelmaddie]

I need help with determining solution.

 

[MarkFL]

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 $c$?

 

[rachelmaddie]

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 $c$?

Are my calculations correct?

 

[MarkFL]

Are my calculations correct?

I think you used:

[MATH] B=\arcsin\left(34\sin\left(\frac{50^{\circ}}{40}\right)\right)[/MATH]
This isn't correct.

 

[rachelmaddie]

I think you used:

[MATH] B=\arcsin\left(34\sin\left(\frac{50^{\circ}}{40}\right)\right)[/MATH]
This isn't correct.

I don’t understand what you did to get 20/17?

 

[MarkFL]

I don’t understand what you did to get 20/17?

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]

 

[rachelmaddie]

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]

So how do I determine the solution?

 

[MarkFL]

So how do I determine the solution?

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 $c$?

 

[rachelmaddie]

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 $c$?

Okay, I’ve never worked with reducing fractions in this case?

 

[MarkFL]

Okay, I’ve never worked with reducing fractions in this case?

It's not necessary to reduce fractions to find the angles, but it's just good form to do so.

 

[rachelmaddie]

It's not necessary to reduce fractions to find the angles, but it's just good form to do so.

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

 

[MarkFL]

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]

 

[rachelmaddie]

34sin(65.7 degrees)/sin(50 degrees) = 33.9
34sin(14.3 degrees)/sin(50 degrees) = 11.0

 

[MarkFL]

As a general rule, I would advise against using the values rounded to one decimal place in intermediary calculations.

 

[rachelmaddie]

As a general rule, I would advise against using the values rounded to one decimal place in intermediary calculations.

Usually they ask to round to the nearest tenth.

 

[MarkFL]

Usually they ask to round to the nearest tenth.

All rounding should be done at the very end, for each calculation.