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Can you solve this triangle?

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  1. Dec 7, 2017 #1
    triangle-base.gif 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. jcsd
  3. Dec 7, 2017 #2

    berkeman

    User Avatar

    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
    triangle-base.gif
     
  4. Dec 7, 2017 #3
  5. Dec 7, 2017 #4

    Mark44

    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##
     
  6. Dec 7, 2017 #5
    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
  7. Dec 7, 2017 #6

    Merlin3189

    User Avatar
    Gold Member

    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
  8. Dec 7, 2017 #7

    Mark44

    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##.
     
  9. Dec 7, 2017 #8
    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.
     
  10. Dec 7, 2017 #9

    Merlin3189

    User Avatar
    Gold Member

    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?
     
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