Geometry determining solutions

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rachelmaddie

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Is my work correct? Also I need a little more understanding of how to determine the solution.
 

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First, let's resize and rotate the image so that it is oriented correctly and display it inline so people can read it as they post:

fmh_0085.jpeg
 
Okay, it's difficult to tell exactly what you are given with which to begin. I am assuming you are given side \(r=11\) and \(\angle P=37^{\circ}\). Is this correct?
 
Okay, as you did, I would begin with the Law of Sine to determine angle \(R\)

[MATH]\frac{\sin(R)}{11}=\frac{\sin\left(37^{\circ}\right)}{9}[/MATH]
[MATH]\sin(R)=\frac{11\sin\left(37^{\circ}\right)}{9}[/MATH]
[MATH]R=\arcsin\left(\frac{11\sin\left(37^{\circ}\right)}{9}\right)\approx47.4^{\circ}\quad\checkmark[/MATH]
But, we must also consider:

[MATH]R=180^{\circ}-\arcsin\left(\frac{11\sin\left(37^{\circ}\right)}{9}\right)\approx132.6^{\circ}[/MATH]
Do you see why?
 
MarkFL said:
Okay, as you did, I would begin with the Law of Sine to determine angle \(R\)

[MATH]\frac{\sin(R)}{11}=\frac{\sin\left(37^{\circ}\right)}{9}[/MATH]
[MATH]\sin(R)=\frac{11\sin\left(37^{\circ}\right)}{9}[/MATH]
[MATH]R=\arcsin\left(\frac{11\sin\left(37^{\circ}\right)}{9}\right)\approx47.4^{\circ}\quad\checkmark[/MATH]
But, we must also consider:

[MATH]R=180^{\circ}-\arcsin\left(\frac{11\sin\left(37^{\circ}\right)}{9}\right)\approx132.6^{\circ}[/MATH]
Do you see why?
Click to expand...
It tells you that the angle is obtuse?
 
rachelmaddie said:
It tells you that the angle is obtuse?
Click to expand...

Well, in one possible solution, we do find that \(R\) is obtuse, but in the SSA case (where we are given 2 sides and an angle) we find this can be the so-called "ambiguous case," that is we may find two distinct triangle that fit the given data. Recall the identity for sine:

[MATH]\sin(180^{\circ}-x)=\sin(x)[/MATH]
Since 47.4 > 37, we know the supplementary angle to \(47.4^{\circ}\) will also work, because this difference allows for a positive value for the third angle in the second case.

Does this make sense?
 
MarkFL said:
Well, in one possible solution, we do find that \(R\) is obtuse, but in the SSA case (where we are given 2 sides and an angle) we find this can be the so-called "ambiguous case," that is we may find two distinct triangle that fit the given data. Recall the identity for sine:

[MATH]\sin(180^{\circ}-x)=\sin(x)[/MATH]
Since 47.4 > 37, we know the supplementary angle to \(47.4^{\circ}\) will also work, because this difference allows for a positive value for the third angle in the second case.

Does this make sense?
Click to expand...
I’m a little confused with this concept.
 
No, this one:

[MATH]\sin(180^{\circ}-x)=\sin(x)[/MATH]
 
It can be derived using the angle difference formula for sine, but the way I visualize it is to picture the unit circle, and a first point at (1,0) and a second point at (-1,0).

Let the first point begin moving at a constant speed around the circle in a counter-clockwise direction, while at the same time the second point begins moving at the same speed but in a clockwise direction.

The \(y\)-coordinate of the first point can represent \(\sin(x)\) while the \(y\)-coordinate of the second point can represent \(\sin(180^{\circ}-x)\).

Can you see that the \(y\)-coordinates of the two points will always be the same?
 
MarkFL said:
It can be derived using the angle difference formula for sine, but the way I visualize it is to picture the unit circle, and a first point at (1,0) and a second point at (-1,0).

Let the first point begin moving at a constant speed around the circle in a counter-clockwise direction, while at the same time the second point begins moving at the same speed but in a clockwise direction.

The \(y\)-coordinate of the first point can represent \(\sin(x)\) while the \(y\)-coordinate of the second point can represent \(\sin(180^{\circ}-x)\).

Can you see that the \(y\)-coordinates of the two points will always be the same?
Click to expand...
I can not say that I am able to visualize that.
 
MarkFL has hinted in post #6 as to whether there are no solutions, one solution or two solutions. You've ruled the first out because you have one solution already. What do you think?
 
Yes, or you can manually drag the slider (the circular button under the t value) left and right.
 
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