Calc the length of side of a triangle given One angle and One Side and ratio x

[John26]

Hi I am posting this again as initially I tried to include the drawing of the triangle but see that it did not attach/show.

Hi, I am a parent trying to help my 15 year old son with the following question,

In a triangle ABC, AB=3x cm, AC=x cm, BC=26 cm and angle BAC = 120 deg

Question: Calculate the value of x given:
angle A = 120 deg
side a (BC) = 26cm
side b (AC) = x cm
side c (AB) = 3x cm

Now I assume you need to use the cosine rule but you are only given the the one length and one angle, I have tried to substituet in "x" and "3x" for b and c into the cosine rule.

a^2 = b^2 + c^2 - 2bc * cosA

b=x
c=3x

26^2 =x^2 + 3x^2 - (2 * x * 3x * cos 120 deg)
676 =x^2 + 9x^2 - (6x^2 * -0.5)
676 = 10x^2 - 3x^2
676 = 7x^2
676/7 = x^2
96.57 = x^2
x = 9.83
therefore side AC (b) = 9.83 cm and side AB (c) = 3*9.83 = 29.48 cm which cannot be correct as it can not be longer than side a (BC 26cm) due to the 120 deg angle.
Can someone help, am I on the right track or is this completely wrong
Any help would be appreciated.
Regards
John

 

[stapel]

In a triangle ABC...calculate the value of x given:
angle A = 120 deg
side a (BC) = 26cm
side b (AC) = x cm
side c (AB) = 3x cm

Now I assume you need to use the cosine rule....

a^2 = b^2 + c^2 - 2bc * cosA

26^2 =x^2 + 3x^2 - (2 * x * 3x * cos 120 deg)
676 =x^2 + 9x^2 - (6x^2 * -0.5)
676 = 10x^2 - 3x^2

Check your signs: -(6x^2)(-0.5) = +3x^2

Everything else was fine, and you should get a better answer after making the above change.

 

[John26]

Thanks

Check your signs: -(6x^2)(-0.5) = +3x^2

Everything else was fine, and you should get a better answer after making the above change.

Stapel, thanks really appreciate your help

Can sleep again