Solving for the third one should be straightforward.

[Tonyb24]

Homework Statement

Completely solve this triangle. No calculators please.
A=?
B=Pi/3
C=?
a=(1+sqrt(3))
b=?
c=2

Homework Equations

Cosine law: b^2=a^2c^2-2ac(cos(B))
Sine law: Sin(A)/a=Sin(B)/b

The Attempt at a Solution

b^2=-6
You can plug in 1/2 in (cos(B)) right away.
Other attemps, don't ask what I did but I ended up finding A=pi/3 and that makes no sense.

 

[berkeman]

Homework Statement

Completely solve this triangle
A=?
B=Pi/3
C=?
a=(1+sqrt(3))
b=?
c=2

Homework Equations

b^2=a^2c^2-2ac(cos(B))
Sin(A)/a=Sin(B)/b

The Attempt at a Solution

b^2=-6
You can plug in 1/2 in (cos(B)) right away.
Other attemps, don't ask what I did but I ended up finding A=pi/3 and that makes no sense.

Welcome to the PF.

Can you show the figure with the triangle? You can Upload a PDF or JPEG file with the button in the lower right...

Is it like this?

https://www.calculatorsoup.com/images/triangletheorems/triangle-base.gif

 

[Tonyb24]

Welcome to the PF.

Can you show the figure with the triangle? You can Upload a PDF or JPEG file with the button in the lower right...

Is it like this?

https://www.calculatorsoup.com/images/triangletheorems/triangle-base.gif
View attachment 216304

Yes it is like that. Thanks.

 

[Mark44]

View attachment 216306

Homework Statement

Completely solve this triangle. No calculators please.
A=?
B=Pi/3
C=?
a=(1+sqrt(3))
b=?
c=2

Homework Equations

Cosine law: b^2=a^2c^2-2ac(cos(B))
Sine law: Sin(A)/a=Sin(B)/b

The Attempt at a Solution

b^2=-6

Obviously, this isn't right. Please show your work leading up to this value. I suspect that you made a mistake in squaring $1 + \sqrt 3$

You can plug in 1/2 in (cos(B)) right away.
Other attemps, don't ask what I did but I ended up finding A=pi/3 and that makes no sense.

 

[Tonyb24]

Obviously, this isn't right. Please show your work leading up to this value. I suspect that you made a mistake in squaring $1 + \sqrt 3$

I squared 1+sqrt(3) as (1+sqrt(3))^2=(1+2sqrt(3)+3) =4+2sqrt(3)
For context, this was on my test. Was wondering if someone could solve this from scratch.
So
plugging everything in

b^2= (4+2sqrt(3))+4-(2(1+sqrt(3))(2)Cos(pi/3)
add 4+4+2sqrt(3)=8+2sqrt(3), and finish the"-2acCos(pi/3)" part
b^2=(8+2sqrt(3))-(2+2sqrt(3)(2)(1/2)
(2)(1/2)=1 so cancel that. simplify..
b^2=6
b=sqrt(6)

Lol ok. I messed up my negatives. This is one of many different forms of answers I got. I also tried factoring the hole thing in many different ways during the test and pluging it into the Sine law, but it was very hard to arcsin these. I do not know how to do arcsin(a+b) It was not in any lecture. And then for angle C starting to stack arcsines was starting to seem a bit ridiculous for some reason.

Now, how to find A without a calculator?
Sin(A)/a=Sin(B)/b

Sin(A)/(1+sqrt(3))=Sin(pi/3)/b
Sin(A)=(1+sqrt(3))((sqrt(3)/2))/sqrt(6))
Sin(A)=(1+sqrt(3))(sqrt(3)/2sqrt(6))
Sin(A)=(1+sqrt(3))(1/2sqrt(2))

So I know 1/sqrt(2)=Sqrt(2)/2 .Anyways, this is what I was thinking on the test and was like dammit...

Sin(A)=((1/2sqrt(2))+(sqrt(3)/2sqrt(2))
Sin(A)= (1+sqrt(3))/2sqrt(2)

So how do I arcsine[(1+sqrt(3))/2sqrt(2)] on paper without a calculator? Maybe try to move things around?
arcsin[1/2sqrt(2)+1/2sqrt(3/2)] errm? What? Can I even simplify this way?

 

[Merlin3189]

I thought the clue was in the "don't use a calculator" bit. The calculations must be easy, or maybe you don't need calculations.

You said yourself,

You can plug in 1/2 in (cos(B)) right away.

So, how did you know that?
Do you know any other angles with easy trig ratios? And how do you know them?

 

[Mark44]

Now, how to find A without a calculator?
Sin(A)/a=Sin(B)/b

I would use sides b and c, rather than sides a and c.
I think that the "don't use a calculalator" means to give the exact values for the angles, rather than the approximations that you get with a calculator. Unless I've made a mistake, C will be the arcsin of a number that involves $\sqrt 6$.

 

[Tonyb24]

I thought the clue was in the "don't use a calculator" bit. The calculations must be easy, or maybe you don't need calculations.

You said yourself,
So, how did you know that?
Do you know any other angles with easy trig ratios?

I know pi pi/2 pi/3 pi/4 and pi/6. I can do some fractions like 7pi/12 using ex.: sin(pi/4+pi/3)=sin(7pi/12)=Sin(pi/4)Cos(pi/3)+Sin(pi/3)Cos(pi/4)
The don't use a calculator bit means exact answers as Mark said.

 

[Merlin3189]

So how do you remember sin(π/3) or cos(π/4) say?
Perhaps you just know them off by heart. In that case how did you find out their exact values, that calculators and tables don't give you?

 

[jimkris69]

An easy thing to do is drop a vertical line from the apex of angle C to the base, c. Now you will have two right triangles facing each other with a common side and you can use simple trig to get the lengths of the various sides. From those lengths , figure out what the angles must be.

 

[LCKurtz]

You have $b$ correct. It's easy to get C using the law of sines with B (have no fear about the $b=\sqrt 6$). Then once you know two angles...