How does A =13.9 degrees,
cosA=16+13-1/2*4*square root of 13
A=cos^-1(16+13-1/2*4*square root of 13)
A=13.9degrees

Solve the triangle.
This is what this problem was part of;
a=1,b=4,C=60degrees
I found C, which was square root of 13

Question

How does A =13.9 degrees,
cosA=16+13-1/2*4*square root of 13
     A=cos^-1(16+13-1/2*4*square root of 13)
     A=13.9degrees

Solve the triangle.
This is what this problem was part of;
a=1,b=4,C=60degrees
I found C, which was square root of 13

raden · Answer

use the cosine rule :
cosA = (b^2 + c^2 - a^2)/(2bc)

raden · Answer

yeah, u have done it.. :)
just use ur calculator to get Cos^-1

anonymous · Answer

Cos^-1 is .5403023059

raden · Answer

cosA = (16+13-1)/(2*4*sqrt(13))
cosA = 0.97
A = cos^-1 (0.97) = 13.86
recheck ur calculating

anonymous · Answer

I get the 0.97 part but when I enter in A = cos^-1 (0.97),
I dont get 13.86?

raden · Answer

but i got :)

anonymous · Answer

you entered in cos^-1 (0.97) in the calcualtor right?

raden · Answer

yep...

anonymous · Answer

when I entered it I get .5240932367?

anonymous · Answer

that was in Ti83,
if I enter it in wolfram site I get 14.07degrees

raden · Answer

hmmm i dont understand use Ti83, i just use from calculator in my laptop. and got it