Question

in triangle abc; a=14cm b=9cm and c=6cm.... solve the triangle..... how do i do this without angle measurements?

Answer 1

if i remember correctly it is often more convenient to solve for the greatest angle first
the greatest angle will be opposite the largest side, so maybe you would be better of starting with
\[\cos(A)=\frac{b^2+c^2-a^2}{2bc}\] we can work through the steps if you like

Answer 2

ok and i made a typo on my very first line
it should be
\[a^2=b^2+c^2-2bc\cos(A)\] or
\[\cos(A)=\frac{b^2+c^2-a^2}{2bc}\]

Answer 3

yes please work through it :)

Answer 4

in your case \(a=14, b=9, c=6\) so you get
\[\cos(A)=\frac{9^2+6^2-14^2}{2\times 9\times 6}=-.731\]rounded

Answer 5

therefore \(A=\cos^{-1}(-.731)=137\) degrees

Answer 6

i computed in one step here
http://www.wolframalpha.com/input/?i=arcos%28%289^2%2B6^2-14^2%29%2F%282*9*6%29%29

Answer 7

now you know 3 sides and one angle, abandon the law of cosines and use the law of sines for the next angle because it is easier

then for the last angle make sure they add up to 180

Answer 8

btw law of cosines is easy to remember if you remember pythagoras, because it looks almost the same
pythagoras for a right triangle \(c^2=a^2+b^2\)
if you do not have a right triangle you need so adjust with
\[c^2=a^2+b^2-2ab\cos(C)\]

Answer 9

oh okay thanks! you have been a huge help!!

Answer 10

yw, you good from here right? with law of sines?

Answer 11

yeah i get the law of sines