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In triangle ABC, we know the following:

 

\(\begin{align*} \angle A &= 120^{\circ} \\ AB &= x \\ AC &= x + 2 \\ BC &= x + 4 \\ \end{align*}\)

 

Find \(x\).

 Jun 3, 2024
 #1
avatar+129487 
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We can use the Law of Cosines

 

BC^2 = AB^2 + AC^2  - 2 ( AB * AC) * cos (120)

 

(x + 4)^2  = x^2  + (x + 2)^2  - 2( x * ( x + 2)) (- 1/2)    simplify

 

x^2 + 8x + 16 =  x^2 + x^2 + 4x + 4  + x^2 + 2x

 

x^2 + 8x + 16  = 3x^2 + 6x  + 4

 

2x^2 - 2x - 12 =  0

 

x^2 -x - 6  =  0

 

(x -3) ( x + 2)   =  0

 

X -3   = 0

 

x = 3

 

 

cool cool cool

 Jun 3, 2024

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