+0  
 
0
53
3
avatar+48 

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
avatar+111330 
+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°

 

cool cool cool

 Mar 30, 2020
 #2
avatar+48 
+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
avatar+111330 
+2

I did  not  notice  that  it was a RT

 

However....the  Law of Cosines  will  always  work

 

 

 

cool cool cool

CPhill  Mar 30, 2020

6 Online Users

avatar