Question

how do i find the interior angles of a triangle with just the side lengths?

Answer 1

You could use law of cosines to get one of the angles and then law of sines to get the rest.

Answer 2

If it is a right triangle, you can use SOH CAH TOA

Answer 3

@dylan0698097 Need more info?

Answer 4

yes i do need more information

Answer 5

Then you need to give me an example.

Answer 6

it isn't a right triangle and side a is 4 side b is 6 and side c is 5

Answer 7

Law of Cosines: \[
c^2=a^2+b^2-ab\cos(C)
\]

Answer 8

i know the formula i looked in my book i just havent learned this and my gf doesnt understand

Answer 9

its an acute triangle

Answer 10

First get \(C\)
Then use
Law of Sines: \[
\frac{\sin(A)}{a}=
\frac{\sin(B)}{b}=
\frac{\sin(C)}{c}
\]

Answer 11

how do i get C

Answer 12

Why don't you plug the sides into the formula and solve for \(C\)?

Answer 13

do i use that formula?

Answer 14

Using the law of cosines you get \(C\)

Answer 15

\(C\) is the angle opposite of the side \(c\).

Answer 16

but in the formula how does it cos(C) is apart of it how do i find that?

Answer 17

\[
\begin{array}{rcl}
c^2 &=& a^2+b^2-2ab\cos C \\
c^2 - a^2 - b^2 &=& -2ab\cos C \\
\frac{c^2 - a^2 - b^2}{-2ab} &=&\cos C \\
\cos^{-1}\left(\frac{c^2 - a^2 - b^2}{-2ab}\right) &=& C \\
\end{array}
\]

Answer 18

All you have to do it put it into the formula.

Answer 19

thank you so much thats helps a lot