Originally Posted by

**Chaim** C----35----A

--\---------/

---\------/

---28--17

-----\--/

------\/

------B

This is what the triangle is suppose to be like.

Law of sine: (sinA/a)=(sinB/b)=(sinC/c)

Law of cosines: a^{2}=b^{2}+c^{2}-2bc(cosA)

b^{2}=a^{2}+c^{2}-2ac(cosB)

c^{2}=a^{2}+b^{2}-2ab(cosC)

Capital letters are the angles, while lowercase letters are the side lengths

a=28, b=35, c=17

Basically I'm trying to **find all the angles of the triangle**.

First I find angle A

28^{2}=35^{2}+17^{2}-2(35)(17)cosA

784=1514-1190cosA

-730=1190cosA

**Angle A is about 52.193 degrees**

Then I find Angle B, but this is where I'm confused at, I try to find it 3 different ways:

1. Using Law of Sine

(sin52.193/28)=(sinB/35)

sinB = 0.988

sin^{-1}(0.988)=**81.115 degrees=Angle B**

2. Using Law of Sine

(sin52.193/28)=(sinC/17)

sinC = 0.479

sin^{-1}(0.479)=28.620

180-52.193-28.620=**99.187=Angle B**

3. Using Law of Cosine

35^{2}=28^{2}+17^{2}-2(28)(27)(cosB)

1225=1073-2(952cosB) mistake in this line ... 2(28)(27) = 952 ... you multiplied by 2 once too many times

152=1904cosB

0.0799=cosB

cos^{-1}(0.0799)=**85.4162=Angle B**

So basically my teacher got somewhere around 99 degrees for angle B, which is close to what I got for solution #2

But I was wondering, __how do you find out which one to use, because they all seem to work__