The sides of a triangle are (x+4),x and (x-4). If the cosine of the largest angle is 1/5 what is x.
This can be done using cosine law.Let the triangle be ABC.
`AB = (x+4)`
`BC = x`
`CA = (x-4)`
In a triangle the largest length of the triangle is opposite to the largest triangle. Largest angle should be at C.
So `(x+4) ` is the largest length and it is opposite to largest angle.
Using cosine law;
`cosC = ((BC)^2+(AC)^2-(AB)^2)/(2BCxxAC)`
`1/5 = (x^2+(x-4)^2-(x+4)^2)/(2xx(x)(x-4))`
`2x(x-4) = 5(x^2+x^2-8x+16-(x^2+8x+16))`
`2x^2-8x = 5x^2-80x`
`3x^2-72x = 0`
`3x(x-24) = 0`
Here the answer should be` x = 0` or `x = 24` . Since `x>0` because of a length the answer is `x = 24` .
So the answer is x = 24.