1. ## Triangle find angles

If $a=7$, $b=4$, $c=9$; Find the angles A, B and C. Give your answer in degrees to at least 3 decimal places.

A=?
B=?
C=?

2. Hello, qbkr21!

Given three sides of a triangle, you need the Law of Cosines.

You should be able to start with: . $a^2\;=\;b^2 + c^2 - 2bc\cos A$

. . and write it in the form: . $\cos A \;= \;\frac{b^2 + c^2 - a^2}{2bc}$

If $a=7,\;b=4,\;c=9$, find the angles $A,\:B,\:C.$

To find angle $A$, use: . $\cos A \:=\:\frac{b^2+c^2-a^2}{2bc}$

. . We have: . $\cos A \;=\;\frac{4^2+9^2-7^2}{2(4)(9)} \;=\;\frac{48}{72}$

. . Hence: . $A \;= \;\cos^{-1}\!\!\left(\frac{48}{72}\right)\;=\;48.1896851$

. . Therefore: . $\boxed{A \:=\:48.190^o}$

To find angle $B$, use: . $\cos B \:=\:\frac{a^2 + c^2 - b^2}{2ac}$

. . We have: . $\cos B\;=\;\frac{7^2 + 9^2 - 4^2}{2(7)(9)} \;=\;\frac{114}{126}$

. . Hence: . $B \;= \;\cos^{-1}\!\!\left(\frac{114}{126}\right) \;=\;25.2087653$

. . Therefore: . $\boxed{B \:=\:25.209^o}$

To find angle $C$, subtract $A$ and $B$ from $180^o.$ **

. . Therefore: . $C \;=\;180^o - 48.190^o - 25.209^o\quad\Rightarrow\quad\boxed{C \:=\:106.601^o}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

** That is, if you trust your answers for $A$ and $B.$

To check, we can find $C$ with: . $\cos C \:=\:\frac{a^2+b^2-c^2}{2ab}$

We have: . $\cos C \;=\;\frac{7^2+4^2-9^2}{2(7)(4)} \;=\;-\frac{16}{56}$

Then: . $C \;=\;\cos^{-1}\!\!\left(-\frac{16}{56}\right)\;=\;106.6015496^o$ . . . close enough!

3. ## Re:

Soroban I really appreciate you help, but I already found the answers, however I need you to work you magic with the word problem I posted just before this problem.