Melody

Ahr yes, that does look erroneous. Thanks for finding it. :)

I will have to go back and see what mistakes led to this one.  Afterall this was just the checking part. 

Lets have a look,

You say mine does not work.

I have said that the sides could be

\(If \;\;x=2+\sqrt6 \\\text{the sides are}\\ 1+\sqrt6,\quad 3+\sqrt6 ,\quad and\quad 3+2\sqrt6\\ \text{I did check this in the same way that I checked your and it also worked.}\)

Let's see if this works

\(\text{My answer works if this statement is true}\\ (3+2\sqrt6)^2=(1+\sqrt6)^2+(3+\sqrt6)^2-2*(1+\sqrt6)(3+\sqrt6)cos120\\~\\ LHS=9+24+12\sqrt6=33+12\sqrt6\\~\\ RHS=1+6+2\sqrt6+9+6+6\sqrt6-2(3+4\sqrt6+6)*-0.5\\ RHS=22+8\sqrt6+(3+4\sqrt6+6)\\ RHS=33+12\sqrt6\\ RHS=LHS\)

I do not know how you got mine to have an angle of 123.59 degrees you did not validate this statement in any way.

The angle in mine is 120 degrees which is the same as the angle is in yours.

So far I think both answers are valid but this does not sit well with me either.

Maybe one of us is wrong but I can see no errors.

Hello guest,

I have already agreed that your answer is correct.

I have shown you that there is a second answer that you have missed.   

   I see that you dispute so I will give you my proof in the next post. (added when I edited))

I do not understand your logic because the cosine rules is

\(a^2=b^2+c^2-2bcCosA\)

Your working does not mention cos120 anyway.

Maybe you have skipped this and said cosA=-1/2 which is certainly true for this problem.

idk because I have not tried super hard to work your logic out.

If you check my answer as I have with yours you will find that it is also a correct answer.

ok guest lets look at your answer to see if it is right.

You said x=4 or x=0

Certainly x canot be 0 becasue then the side lengths will be negative.

If x=4 then the sides will be

3,5 and 7

I'll use the cos rule to see if this works.

\(LHS=7^2=49\\ RHS=9+25-2*3*5cos120\\ RHS=34-30(-cos60)\\ RHS=34+30(0.5)\\ RHS=49\\ RHS=LHS \quad that\; is \;great.\\\)

So your answer of x=4 is correct

But I do not understand where this came from either.

I did it and got completely answers.

\(x=2\pm\sqrt6\)

The minus one is not good becasue 2-sqrt6 <0

\(If \;\;x=2+\sqrt6 \\\text{the sides are}\\ 1+\sqrt6,\quad 3+\sqrt6 ,\quad and\quad 3+2\sqrt6\\ \text{I did check this in the same way that I checked your and it also worked.}\)

So now I am really confused! 

(I would not be so confused if I got your answer as well but I did not get your answer and you did not get mine)

I drew both answers to scale and they both work fine.   

So I am still confused by how guest found only one answer and I found only one different  answer and both answers appear to be correct.

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AREA

Another way to work out area once you have the sides is to use the    Area=absinC     formula.

There must be a relatively simple way to do this but I am out of ideas.

Can another mathematician help?

inverse tan is the same as arctan.   Use atan    

+8 

Which of the following points is on a circle if its center is (-13,-12) and a point on the circumference is (-17, -12)?

use the distance formula to find the distance between those 2 points, that is the radius.  Call it R

The formula for the circle witll be

\((x+13)^2+(y+12)^2=R^2\)

Plug the points in and see which one makes the formula true.

See if you ideas work by plotting you graph in here

Learn from the answers that i have already given you.

Thanks Tertre.