# Finding angles in a triangle.

Printable View

• Feb 5th 2010, 05:04 AM
s2951
Finding angles in a triangle.
a=5m , b=8m , c=7m

a^2= b^2+c^2-2bc Cos A
(5)^2= (8)^2+(7)^2-2(8)(7) Cos A
Cos A= 5^2-8^2+7^2/-2(8)(7)
Cos A= 10/-112
Angle A= -0.0898
Angle A= 95 Degrees 9 minutes

How to find other B & C angle?

This is what i work out but stuck.........

a/Sin A = b/SinB
5/Sin 95 Degrees 9 minutes = 8/Sin B
Sin B = (8)(Sin 95 Degrees 9 minutes)/5
Sin B = 1.5939

Problem is i calculate using sin^-1...

Thanks much
• Feb 5th 2010, 05:08 AM
Solarmew
the same way? (Nod)

b^2= a^2+c^2-2ac Cos B
c^2= b^2+a^2-2ba Cos C

you got angle A wrong too, check your calculations
Cos A= (5^2-8^2-7^2)/(-2(8)(7))
• Feb 5th 2010, 05:10 AM
s2951
Quote:

Originally Posted by Solarmew
the same way? (Nod)

u mean now need to calculate using Cos B using cosine rule?

SOLVED. THX
• Feb 5th 2010, 05:16 AM
s2951
Quote:

Originally Posted by Solarmew
the same way? (Nod)

b^2= a^2+c^2-2ac Cos B
c^2= b^2+a^2-2ba Cos C

you got angle A wrong too

which part?
• Feb 5th 2010, 05:17 AM
Solarmew
either cosine or sine law would've worked :]
good job (Smile)
• Feb 5th 2010, 05:23 AM
Solarmew
minus 7 not plus :] and dun forget the order of operations
• Feb 5th 2010, 05:30 AM
s2951
Quote:

Originally Posted by Solarmew
minus 7 not plus :] and dun forget the order of operations

Noted.