+0

# Trigonometry

0
53
3

Side a=16, side b=6, and side c=17.1

I figured out that angle A is 20.6, but when I try to find the next angle (angle B) I always come just a few decimal points off

What am I doing wrong

this is what I did

Cos(A)=6/17.1

Cos(A)=0.350877193

Cos⁻¹0.350877193=69.45902246

69.45902246-->69.5

My teacher says that the answer is 69.3

So could you tell me why I am off by so little

Mar 30, 2020

#1
+3

Law  of Cosines

6^2  = 16^2   +  17.1^2  -  2 [ 16 * 17.1] cos B       rearrange  as

[ 6^2  - 16^2  - 17.1^2 ]  / [ -2 * 16 * 17.1 ]   = cos B

Take the  cosine inverse

arccos   ( [ 6^2  - 16^2  - 17.1^2 ]  / [ -2 * 16 * 17.1 ]  )  = B ≈ 20.54°

Likewise

[  16^2  -6^2 - 17.1^2 ]  / [ -2 * 6 * 17.1 ]   =  cos  A

arccos  ( [  16^2  -6^2 - 17.1^2 ]  / [ -2 * 6 * 17.1 ] )  =  A ≈   69.34°   Mar 30, 2020
#2
+3

why do you need to use law of cosines?

The triangle is a right triangle, so can´t you use the normal cos(A)= adjacent over hypotenuse

gvcarden  Mar 30, 2020
#3
+2

I did  not  notice  that  it was a RT

However....the  Law of Cosines  will  always  work   CPhill  Mar 30, 2020