+845 
i tried using the cosine rule for the angle for this one
cos120 = (x-1)^2+(x+1)^2-(2x-1)^2 / 2(x-1)(x+1)
then i managed to simplify it down to cos120 = -2x^2 +4x+1/ 2
but i cant do it further nor do i know how to find x at this step so i believe my approach was completely wrong
please help thanks
Use the Law of Cosines to solve for x:
2 x (x - 2) = (x - 1) (x + 1) + 1
Expand out terms of the left hand side:
2 x^2 - 4 x = (x - 1) (x + 1) + 1
Expand out terms of the right hand side:
2 x^2 - 4 x = x^2
Subtract x^2 from both sides:
x^2 - 4 x = 0
Factor x from the left hand side:
x (x - 4) = 0
Split into two equations:
x - 4 = 0 or x = 0
Add 4 to both sides:
x = 4 or x = 0
Sides: (2*4 - 1)=7, (4 + 1)=5, (4 - 1)=3
Use Heron's formula to calculate the area:
Area =6.495
Used the Law of Cosines to solve for x, for which I got x = 4. [This is what you should get if you don't make mistakes].
Then solve for sides:7, 5, 3.
Once you have the 3 sides, then I used Heron's formula to calculate the area =6.495.
+118687 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.
----------------
AREA
Another way to work out area once you have the sides is to use the Area=absinC formula.
Solve in 2 steps as follows:
Step 1:
Expand the following:
(2 x - 1)^2 = (x + 1)^2 + (x - 1)^2
(x + 1) (x + 1) = (x) (x) + (x) (1) + (1) (x) + (1) (1) = x^2 + x + x + 1 = x^2 + 2 x + 1:
(2 x - 1)^2 = x^2 + 2 x + 1 + (x - 1)^2
(x - 1) (x - 1) = (x) (x) + (x) (-1) + (-1) (x) + (-1) (-1) = x^2 - x - x + 1 = x^2 - 2 x + 1:
(2 x - 1)^2 = x^2 - 2 x + 1 + x^2 + 2 x + 1
(2 x - 1) (2 x - 1) = (2 x) (2 x) + (2 x) (-1) + (-1) (2 x) + (-1) (-1) = 4 x^2 - 2 x - 2 x + 1 = 4 x^2 - 4 x + 1:
4 x^2 - 4 x + 1 = x^2 + x^2 + 2 x - 2 x + 1 + 1
Grouping like terms, x^2 + x^2 + 2 x - 2 x + 1 + 1 = (x^2 + x^2) + (2 x - 2 x) + (1 + 1):
4 x^2 - 4 x + 1 = (x^2 + x^2) + (2 x - 2 x) + (1 + 1)
x^2 + x^2 = 2 x^2:
4 x^2 - 4 x + 1 = 2 x^2 + (2 x - 2 x) + (1 + 1)
1 + 1 = 2:
4 x^2 - 4 x + 1 = 2 x^2 + (2 x - 2 x) + 2
2 x - 2 x = 0:
4 x^2 - 4 x + 1 = 2 x^2 + 2
Subtract 2 x^2 + 2 from both sides of 4 x^2 - 4 x + 1 = 2 x^2 + 2:
-2 x^2 + 2 + 4 x^2 - 4 x + 1 = (2 x^2 + 2) - 2 x^2 + 2
(2 x^2 + 2) - (2 x^2 + 2) = 0:
-(2 x^2 + 2) + 4 x^2 - 4 x + 1 = 0
-(2 x^2 + 2) = -2 - 2 x^2:
-2 - 2 x^2 + 4 x^2 - 4 x + 1 = 0
Grouping like terms, 4 x^2 - 2 x^2 - 4 x - 2 + 1 = (4 x^2 - 2 x^2) - 4 x + (1 - 2):
(4 x^2 - 2 x^2) - 4 x + (1 - 2) = 0
4 x^2 - 2 x^2 = 2 x^2:
2 x^2 - 4 x + (1 - 2) = 0
2 x^2 - 4 x + -1 = 0 - This can be simplified further to:
2 (x - 2) x = 1..............(1)
Step 2:
From (1) above, you will subtract: 2(x+1)(x-1)cos(120)
2(x - 2)x =1 - (1 - x^2), solve for x
x = 4 and x=0
Note: with the sides you calculated, the angles are as follows:
21.333, 35.077, 123.59
With the sides I calculated, the angles are as follows:
120, 21.787, 38.213
+118687 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.
+118687 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.
I used the Law of Cosines exactly as you suggested:
a^2=b^2+c^2-2bcCosA
(2x -1)^2 =[(x+1)^2 + (x-1)^2] - 2*(x+1)(x-1)*cos(120), solve for x
(2 x - 1)^2 = (x - 1)^2 - (1 - x) (x + 1) + (x + 1)^2
Expand out terms of the right hand side:
(2 x - 1)^2 = 3 x^2 + 1
Subtract 3 x^2 + 1 from both sides:
-1 - 3 x^2 + (2 x - 1)^2 = 0
Expand out terms of the left hand side:
x^2 - 4 x = 0
Factor x from the left hand side:
x (x - 4) = 0
Split into two equations:
x - 4 = 0 or x = 0
Add 4 to both sides:
x = 4 or x = 0
Note: I used this online calculator to verify the angles using "SSS" triangle:
https://www.triangle-calculator.com/?what=sss&a=7&b=5&c=3&submit=Solve
