Question

In triangle ABC, a = 2x, b = 3x + 2, c = radical(12), and angle C = 60. Find x.

Answer 1

I know that I have to use the law of cosines, but I keep getting a different answer from the book.......

Answer 2

Can I see your work?

Answer 3

I just plugged in the sides and angle in the following formula
\[a^2 = b^2 + c^2 - 2bc \cos A\]

Answer 4

Right formula but what did you plug into the letters?

Answer 5

\[(\sqrt{12})^2 = (2x)^2 + (3x+2)^2 - 2(2x)(3x+2) * \cos 60\]

Answer 6

Wrong formula actually. Cos 60 is c's NOT a's

Answer 7

That's the general formula

Answer 8

You can flip the formula around

Answer 9

IK but you have to flip it around first and then solve for C^2

Answer 10

and then you use sine formula to make life easier

Answer 11

By flip around, I mean you can switch the letters. Instead of a, you can b or c. It's just a general formula that can be changed around to best suit the problem.

Answer 12

IKIKIKIKIKIK

Answer 13

Did you do that doe? And you still didn't get the right answer? Did you check the textbook's work? You know what let me work it out on a piece of paper and I will ask you what you got okay?

Answer 14

if we got the same answer then the book is wrong okay?

Answer 15

I did do that.
The textbook did it the same way I did it, but somehow got the wrong answer.
I can type up the textbook's explanation if you want?

Answer 16

Yes please? I am so sorry for bombarding you with tonz of questions dude.

Answer 17

@Albert0898 I will be back in 30 minutes or so cause I have to drive back home so I will give you your answer as soon as i get home so sorry!!!

Answer 18

It's ok haha

Here's the textbook's explanation.
---------

Law of Cosines:
\[12 = (2x)^2 + (3x+2)^2 - 2(2x)(3x+2) \cos60\]
\[12 = 4x^2 + 9x^2 + 12x + 4 - (12x^2 + 8x) \frac{ 1 }{ 2 }\]
\[7x^2 + 8x - 8 = 0\]

Use program QUADFORM to get \[x = pm0.64\]
Since a side of a triangle must be positive, x can only equal 0.64.

Answer 19

yeah thats correct and this is the solution that u'll be getting by solving your equation
u might be doing some calculation mistake
check again :)

Answer 20

Okay @imqwerty

In this book, it does not divide both sides by cos 60.
That's where my mistake was.

In my ACT Textbooks, it would say that I would have to do that.

I'm very confused.

I see my mistake now, but am still very confused as to why two different books have two different methods that yield two very different answers.

Answer 21

:) we know that cos 60 is \(\large \frac{1}{2}\) so just simply put it there in the formula and rest of simplification is normal.

Answer 22

But why is it that when I divide 1/2 from both sides, it doesn't work out to get the right answer?

Answer 23

it must work alright
can you show your work?

Answer 24

Sure thing!
\[12 = (4x)^2 + (9x)^2 + 12x + 4 - (12x^2 + 8x)(\frac{ 1 }{ 2 })\]

\[24 = 13x^2 + 12x + 4 - (12x^2 + 8x)\]

\[24 = x^2 + 4x + 4\]

\[x^2 + 4x - 20 = 0\]

Answer 25

1st step has a lil mistake->

\(12 = (4x^{\color{red}2})^{\color{red}1} + (9x^{\color{red}2})^{\color{red}1} + 12x + 4 - (12x^2 + 8x)(\frac{ 1 }{ 2 })\)

and then when u divide by \(\large \frac{1}{2}\) it goes like this-

\(\large\frac{12}{\frac{1}{2}} = \frac{13x^2+12x + 4 - (12x^2 + 8x)(\frac{ 1 }{ 2 })}{\frac{1}{2}}\)

\(24 = \large{\frac{13x^2 + 12x + 4}{\color{red}{\frac{1}{2}}}} - (12x^2 + 8x)\)

did you see the fault? :)

Answer 26

Yes, I did think of that as my fault.
However, my ACT books (which I don't use anymore) wouldn't state that as a method.
It might have been a typo though

Answer 27

its good to directly substitute the value rather than dividing because dividing will make the process lengthy in most cases

Answer 28

Okay, well I learned something new today!
Thank you @imqwerty ! :)

Answer 29

np :)

Answer 30

welp nvm den sorry!