Calculating possible angles of triangle (difficult?)

[TriangleTrouble]

Hello, we have seem to run into some trouble when trying to see what angles would work with a certain triangle.

Information:

sides:

AB = 2
BC = 3
angle A (CAB)= 1/6 pi


So first we wanted to know the lenght of AC, with cosine rule this gives:

(BC)² = (AB)² + (AC)² - 2*AB*AC*cos (angle A)

This gives (AC)² - 2 * √3 * (AC) - 5 = 0

So with quadratic equation this gives AC = √3 + 2*√2 or AC = √3 - 2*√2. Because the last one is negative this is discarded, because lengths can't be negative.


So now, with the Sine rule we then get:


BC / sine(1/6 pi) = AC / sine (angle ABC).

This gives a angle of 49 degrees or 131 degrees (because of the unit circle with sinus and having a positive outcome).

HOWEVER when you try to draw the actual trying it seems the angle of 49 degrees is impossible.


So basically the question is:

How can we see that that angle is impossible without drawing it or did we do something wrong and should it actually be possible?
I hope this was clear, thanks in advance.

 

[Deleted member 4993]

Hello, we have seem to run into some trouble when trying to see what angles would work with a certain triangle.

Information:

sides:

AB = 2
BC = 3
angle A (CAB)= 1/6 pi


So first we wanted to know the lenght of AC, with cosine rule this gives:

(BC)² = (AB)² + (AC)² - 2*AB*AC*cos (angle A)

This gives (AC)² - 2 * √3 * (AC) - 5 = 0

So with quadratic equation this gives AC = √3 + 2*√2 or AC = √3 - 2*√2. Because the last one is negative this is discarded, because lengths can't be negative.


So now, with the Sine rule we then get:


BC / sine(1/6 pi) = AC / sine (angle ABC).

This gives a angle of 49 degrees or 131 degrees (because of the unit circle with sinus and having a positive outcome).

HOWEVER when you try to draw the actual trying it seems the angle of 49 degrees is impossible.


So basically the question is:

How can we see that that angle is impossible without drawing it or did we do something wrong and should it actually be possible?
I hope this was clear, thanks in advance.

Calculate the length of AC - applying law of cosine

 

[Ishuda]

...

So basically the question is:

How can we see that that angle is impossible without drawing it or did we do something wrong and should it actually be possible?
I hope this was clear, thanks in advance.

Also use the Law of Sines for angle C and you will get two angles [19\(\displaystyle ^\circ\) and 161\(\displaystyle ^\circ\)]. Now the angles B [either 131\(\displaystyle ^\circ\) or 49\(\displaystyle ^\circ\)] and C have to add to 150\(\displaystyle ^\circ\) [since angle A is 30\(\displaystyle ^\circ\)]. There are four combinations, only of which add to 150\(\displaystyle ^\circ\).