College Level Beginning Trigonometry, Solving Triangles

[aexel]

Hello, in my trigonometry class we learned how to solve for triangles. It involves finding the missing sides and angles when given 3 pieces of information. I can figure them out easily when I am given two angles, or two sides with the angle in between, but this one has me very confused.

My homework problem is: <B = 73.4, a=859, b=783

So i began by drawing and labeling my triangle. I think it is labeled SSA which is the ambiguous case because <B is no in-between the two sides. According to my notes I can have either no triangle, 1 triangle, or 2 triangles forming with the given measurements. So I drew a line down the middle of the triangle and labeled it h because that determines what triangle scenario I will have.To find h, my notes say "h=b*sinA". I then attempted to find SinA by using law of sines.

SinA / a = Sin B / b Sin A = Sin74.3 *859 / 783

SinA= Sin 73.4 * 859 Sin A = 1.04999 Then I applied the sin inverse and got A = 0.01832.


A=0.01832 is what is causing my confusion. That angle is so small compared to <B. I don't know what I did wrong, and yes my calculator is in degree mode.

So when I want to calculate h. I get H= 783* sin 0.01832 H= 0.250

Comparing a and h. I have 859> 0.250, therefore there are two triangles? I'm completely lost here because I showed this to a tutor and he simply said that there is no triangle because of the value of theta. I don't know how he came up with that. He is in advanced math, so it made me more confused.

All help is highly appreciated.

 

[Deleted member 4993]

Hello, in my trigonometry class we learned how to solve for triangles. It involves finding the missing sides and angles when given 3 pieces of information. I can figure them out easily when I am given two angles, or two sides with the angle in between, but this one has me very confused.

My homework problem is: <B = 73.4, a=859, b=783

So i began by drawing and labeling my triangle. I think it is labeled SSA which is the ambiguous case because <B is no in-between the two sides. According to my notes I can have either no triangle, 1 triangle, or 2 triangles forming with the given measurements. So I drew a line down the middle of the triangle and labeled it h because that determines what triangle scenario I will have.To find h, my notes say "h=b*sinA". I then attempted to find SinA by using law of sines.

SinA / a = Sin B / b Sin A = Sin74.3 *859 / 783

SinA= Sin 73.4 * 859 Sin A = 1.04999 Then I applied the sin inverse and got A = 0.01832.


A=0.01832 is what is causing my confusion. That angle is so small compared to <B. I don't know what I did wrong, and yes my calculator is in degree mode.

So when I want to calculate h. I get H= 783* sin 0.01832 H= 0.250

Comparing a and h. I have 859> 0.250, therefore there are two triangles? I'm completely lost here because I showed this to a tutor and he simply said that there is no triangle because of the value of theta. I don't know how he came up with that. He is in advanced math, so it made me more confused.

All help is highly appreciated.

The given parameters cannot form a triangle because |sin(Θ)| ≤ 1

So sinA = 1.05 is not POSSIBLE.

 

[aexel]

The given parameters cannot form a triangle because |sin(Θ)| ≤ 1

So sinA = 1.05 is not POSSIBLE.

So was my math for Sin A correct? Was i suppose to apply the inverse or by seeing that Sin A = 1.05 was it? We weren't taught that sin theta < or = to 1. Is there another way I can see that? I think this is what the tutor was talking about. But in terms of h > A how would I come up with no triangle?

 

[Deleted member 4993]

So was my math for Sin A correct? Was i suppose to apply the inverse or by seeing that Sin A = 1.05 was it? We weren't taught that sin theta < or = to 1. Is there another way I can see that? I think this is what the tutor was talking about. But in terms of h > A how would I come up with no triangle?

That may not have been taught explicitly - but you should know it as soon as SINE of an angle is defined to you.

$\displaystyle Sin(\Theta) \ = \ \dfrac{side}{hypotenuse}$

length of any side of a right-triangle is always ≤ hypotenuse hence |sin(Θ)| ≤ 1

 

[aexel]

Not sure what you're doing...like, how can a triangle be labelled SSA

A triangle is USUALLY labelled ABC, angles being A, B anc C,
side opposite A labelled a, opposite B labelled b and opposite C labelled c,
with a < b < c; like:

          C


 

B                        A

A = ?, B = ?, C = 73.4, a = ?, b = 783, c = 859


Using Law of Sines, you'll get A = ~45.8, B = ~60.8 and a = ~642

Are you able to get those results?

If so, then what exactly is your problem?

I should have said I classified it as SSA. It is labeled as you said, but c is not equal to 859. a =859. In my calculations I get an error, but I see that it is because Sin A = 1.05 which is larger than 1. Therefore, it does not exist.

 

[aexel]

That may not have been taught explicitly - but you should know it as soon as SINE of an angle is defined to you.

$\displaystyle Sin(\Theta) \ = \ \dfrac{side}{hypotenuse}$

length of any side of a right-triangle is always ≤ hypotenuse hence |sin(Θ)| ≤ 1

Oh ok, I see now. I also calculated h wrong. That would have helped me see that no triangle can exist. Thank you.