Solving Triangle ABC: Finding Value of x

[Natasha1]

Homework Statement

In triabgle ABC, AB = 2x cm, AC = x cm, BC = 21 cm and angle BAC = 120 degrees. Calculate the value of x?

Homework Equations

I used the cosine rule here

The Attempt at a Solution

As we have 2 sides and an angle I used the cosine rule - Am I correct?
(2) means squared

b(2) = a(2) + c(2) - 2ac cos B
so
x(2) =21(2) + (2x)(2) - 2.21.2x.c cos B
x(2) =441 + 4x(2) - 84x cos B
3x(2) = 84x cosB - 441

How do I find x from here?

 

[Merlin3189]

Why did you choose to use angle B ?
Since you know angle A, wouldn't that be easier?

 

[Samy_A]

Homework Statement

In triabgle ABC, AB = 2x cm, AC = x cm, BC = 21 cm and angle BAC = 120 degrees. Calculate the value of x?

Homework Equations

I used the cosine rule here

The Attempt at a Solution

As we have 2 sides and an angle I used the cosine rule - Am I correct?
(2) means squared

b(2) = a(2) + c(2) - 2ac cos B
so
x(2) =21(2) + (2x)(2) - 2.21.2x.c cos B
x(2) =441 + 4x(2) - 84x cos B
3x(2) = 84x cosB - 441

How do I find x from here?

Applying the cosine rule is the way to go.
But you should apply it correctly, with the angle you know (that's angle A):
$a²=b²+c²-2bc\cos(A)$, where a is the side opposite angle A, b the side opposite angle B and c the side opposite angle C.
You also know $\cos(A)=\cos(120°)$.

 

[Natasha1]

Thanks everyone, I have found that x(2) = 63 which makes x = 7.94 cm (to 2dp) using the cosine rule

Am I correct?

a(2) = b(2) + c(2) - 2bc cos B
so
21(2) =x(2) + (2x)(2) - 2.x.2x cos B
441 = 5x(2) - 4x(2) cos 120
441 = x(2) (5-4 cos 120)
x(2) = 441 / (5-4cos 120)
x(2) = 63
x = 7.94 cm (to 2dp)

 

[Samy_A]

Thanks everyone, I have found that x(2) = 63 which makes x = 7.94 cm (to 2dp) using the cosine rule

Am I correct?

a(2) = b(2) + c(2) - 2bc cos B
so
21(2) =x(2) + (2x)(2) - 2.x.2x cos B
441 = 5x(2) - 4x(2) cos 120
441 = x(2) (5-4 cos 120)
x(2) = 441 / (5-4cos 120)
x(2) = 63
x = 7.94 cm (to 2dp)

Looks correct (except that the angle should better be called A, not B).