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

YEEEEEET Jan 26, 2019

#1**+2 **

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**

Guest Jan 27, 2019

#3**0 **

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.

Guest Jan 27, 2019

#4**+2 **

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.

Melody Jan 27, 2019

#5**0 **

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**

Guest Jan 27, 2019

edited by
Guest
Jan 27, 2019

#6**+1 **

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.

Melody
Jan 27, 2019

#7**+1 **

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.

Melody
Jan 27, 2019

#8**+1 **

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**

Guest Jan 27, 2019