Question

Let ABC be a triangle such that angle acb=pie/6 and let a ,b and c denotes the length of sides opposite to A,B and C respectively .the value of x for which a=x^2+x+1,b=x^2-1 and c=2x+1 is /are?

Answer 1

@perl

Answer 2

@dan815

Answer 3

angle acb is pie/6

Answer 4

woops

Answer 5

not angle abc

Answer 6

let me fix that :)

Answer 7

lol.ok

Answer 8

directions, solve for x?

Answer 9

yes we want values of x .but how ?

Answer 10

i can do this,

we can use herons formula and area of triangle

Answer 11

1/2 ( x^2 + x + 1) * (x^2 -1 ) * sin(pi/6 ) = sqrt ( s( s - a) ( s- b) ( s - c)

Answer 12

one moment

Answer 13

ok

Answer 14

s=( a + b + c ) / 2

Answer 15

hmmm.then

Answer 16

one minute, a lot of algebra here

Answer 17

ok

Answer 18

x = 1 + sqrt(3)

Answer 19

the calculation, equate areas using trigonometry approach and heron's formula.
$$
\large \rm {
Let \\
\\
\\a= x^2+x+1
\\b=x^2-1
\\c=2x+1
\\s =\frac{a+b+c}{2}

\\ \therefore \\
s = x^2+\frac32x+\frac 1 2 ~\\
\\~\\
Equate ~ Area \\
\frac 12 ( x^2 + x + 1) (x^2 -1 )\sin(\pi/6 ) = \sqrt{ s( s-a)(s-b)(s-c)}


}
$$

Answer 20

$$ \Large{
\rm x = 1 + \sqrt 3 ,~ -2 - \sqrt 3, ~ 1, ~ -1 \\

\\~\\ ignore~ x = 1 ,~ and ~the ~negative ~solutions
}
$$

Answer 21

because we assume length has to be positive, if you plug in x = 1, you get zero, x = -1 you get zero. i didnt try the last one , one moment

Answer 22

yes, if you plug -2-sqrt(3) you get a negative length for side c

Answer 23

I can't show you my algebra because it is pages long

Answer 24

how did you do it?

Answer 25

ohh.ok.thanks

Answer 26

this definitely works, its a lot of work to do it by hand . make clever substitutions along the way

Answer 27

@danish071996