Question

In triangle ABC, AB=2x, AC=x, BC=21 and angle BAC=120. Calculate the value of x. I think i have to use cosine rule but dont get it will fan and medal

Answer 1

@paki

Answer 2

law of cosines

Answer 3

i know so what i did was 21^2= x^2+4x^2-(2 times 2x times x cos 120)? is this right?

Answer 4

draw the triangle, it will be easier to see and solve

Answer 5

so \[21^2=x^2+\left( 2x \right)^2-2\left( x \right)\left( 2x \right)\cos120^\circ\]

Answer 6

yah i drew it yah

Answer 7

from this you should get a quadratic in x. solve for x

Answer 8

then is it x^2 cos 120=21^2?

Answer 9

\(441=5x^2-4x^2(-1/2)\) yeah?

Answer 10

oh ok

Answer 11

so wait

Answer 12

what does this give me?

Answer 13

you tell me... work it out

Answer 14

ah 7x^2=21^2? right

Answer 15

so x would be 7.94...

Answer 16

so \(x^2 = 63 \Rightarrow x = \sqrt{63}\)

Answer 17

good to go?

Answer 18

yup