# Can you solve this triangle?

Tags:
1. Dec 7, 2017

### Tonyb24

1. The problem statement, all variables and given/known data
Completely solve this triangle. No calculators please.
A=?
B=Pi/3
C=?
a=(1+sqrt(3))
b=?
c=2
2. Relevant equations
Cosine law: b^2=a^2c^2-2ac(cos(B))
Sine law: Sin(A)/a=Sin(B)/b
3. The attempt at a solution
b^2=-6
You can plug in 1/2 in (cos(B)) right away.
Other attemps, dont ask what I did but I ended up finding A=pi/3 and that makes no sense.

Last edited: Dec 7, 2017
2. Dec 7, 2017

### Staff: Mentor

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

3. Dec 7, 2017

4. Dec 7, 2017

### Staff: Mentor

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

5. Dec 7, 2017

### Tonyb24

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?

Last edited: Dec 7, 2017
6. Dec 7, 2017

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

Last edited: Dec 7, 2017
7. Dec 7, 2017

### Staff: Mentor

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 $\sqrt 6$.

8. Dec 7, 2017

### Tonyb24

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 dont use a calculator bit means exact answers as Mark said.

9. Dec 7, 2017

### 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?

10. Dec 31, 2017

### 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.

11. Dec 31, 2017

### 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...