# Finding sides of triangles

I have a triangle with

Side A = 2 Side B = 6

Angle c (angle opposite Side C) = 105°

I want to find side C

My first though was to use the SOH CAH TOA rule

In this case we have the adjacent and hypotenuse of angle c so I use CAH (cosθ = A/H)

Because 105° is in the 2nd quadrant it will be -cos

So -cos 105° = 2/C

Therefore C = 2/-cos 105°

        = 7.73


I thought this was the correct way to do it because that's all we have been learning in maths.

But in physics we are taught to use C = √(A²+B² - 2ABcosθ) which gives a different answer.

So which do I use/which is more accurate?

• First rule doesn't apply. This is not a right triangle. Chris is correct. – JoeTaxpayer Aug 29 '13 at 2:17
• Use the law of cosines... SOH CAH TOA requires a right triangle – Eleven-Eleven Aug 29 '13 at 2:19

Use the Cosine Law: $$c^2=a^2+b^2-2ab\cos(\angle C).\tag{1}$$ You know that $a=6$, so $b=3$, so now we know all the items on the right-hand side of (1).
Remark: The SOH CAH TOA stuff is for right-angled triangles. Our triangle is definitely not right-angled, since one of its angles is $105^\circ$.
Note that the Cosine Law is a generalization of the Pythagorean Theorem. If $\angle C=90^\circ$, then $\cos(\angle C)=0$, and the Cosine Law (1) becomes the familiar $c^2=a^2+b^2$.
Your angle is $105$ so the sohcahtoa rule doesn't apply. The law of cosines or the law of sines are for all triangles.