Question

SAFind the value of x.Round to the nearest tenth.X29B50x = [? ]°26C

Answer 1

Law of cosines:

`c^2=a^2+b^2-2ab\cos C`

For the given triangle:

`\begin{gathered} a=26 \\ b=50 \\ c=29 \\ A=x \\ \\ a^2=b^2+c^2-2bc\cos A \end{gathered}`
`26^2=50^2+29^2-2(50)(29)\cos x`

Solve x from the equation above:

`\begin{gathered} 676=2500+841-2900\cos x \\ 676=3341-2900\cos x \\ 676-3341=-2900\cos x \\ -2665=-2900\cos x \\ (-2665)/(-2900)=\cos x \\ \\ \cos x=(533)/(580) \\ \\ x=\cos^(-1)((533)/(580)) \\ \\ x\approx23.2 \end{gathered}`Then, the value of x is 23.2º