# Thread: Law of Sine and Cosine Question

1. ## Law of Sine and Cosine Question

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: a2=b2+c2-2bc(cosA)
b2=a2+c2-2ac(cosB)
c2=a2+b2-2ab(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
282=352+172-2(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
352=282+172-2(28)(27)(cosB)
1225=1073-2(952cosB)
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
Thanks!

2. ## Re: Law of Sine and Cosine Question

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: a2=b2+c2-2bc(cosA)
b2=a2+c2-2ac(cosB)
c2=a2+b2-2ab(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
282=352+172-2(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
352=282+172-2(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
as far as your two solutions for angle B, note that if $b^2 > a^2 + c^2$ , then angle B is obtuse

3. ## Re: Law of Sine and Cosine Question

Also cosB is negative ( -152/952) so B is obtuse as you say (B=99.2)