Question

Triangle ABC has AC = 8x-3, BC = 4x-1, angle ABC = 120 and angle ACB = 15. show that the exact value of x is (9+sqrt6) divided by 20?

I know I have to use the sin rule but I'm not sure where to go from there

Answer 1

See below

Explanation:

The sine rule says

`(AC)/(sin/_ABC) = (BC)/(sin/_BAC)`

Since `/_BAC = 180^circ - /_ABC-/_ACB = 45^circ`, we have

`(8x-3)/(sin120^circ) = (4x-1)/(sin45^circ)qquad implies`

`(8x-3)2/sqrt 3 = (4x-1)sqrt 2implies`

`16x-6 = (4x-1)sqrt6 implies`

`4(4-sqrt6)x = 6-sqrt6`

`x = 1/4 (6-sqrt 6)/(4-sqrt6) = 1/4 (6-sqrt 6)/(4-sqrt6) times (4+sqrt 6)/(4+sqrt6)`

`quad = 1/4 (18+2sqrt6)/(16-6) =(18+2sqrt6)/40 = (9+sqrt6)/20`