Question

Solve VABC if a = 34 feet, b = 20 feet, and c = 18 feet. .

Answer 1

Cosine theorem:

`\begin{gathered} a^2=b^2+c^2-2bccosA \\ b^2=a^2+c^2-2ac\cos B \\ c^2=a^2+b^2-2ab\cos C \end{gathered}`

a= 34ft

b = 20ft

c = 18ft

`\begin{gathered} a^2-b^2-c^2=-2bc\cos A \\ (a^2-b^2-c^2)/(-2bc)=\cos A \\ \\ A=\cos ^(-1)((a^2-b^2-c^2)/(-2bc)) \end{gathered}`
`B=\cos ^(-1)((b^2-a^2-c^2)/(-2ac))`
`C=\cos ^(-1)((c^2-a^2-b^2)/(-2ab))`
`\begin{gathered} A=\cos ^(-1)((34^2-20^2-18^2)/(-2(20)(18))) \\ \\ A=\cos ^(-1)((432)/(-720))=126.86 \end{gathered}`
`\begin{gathered} B=\cos ^(-1)((20^2-34^2-18^2)/(-2(34)(18))) \\ \\ B=\cos ^(-1)((-1080)/(-1224))=28.07 \end{gathered}`
`\begin{gathered} C=\cos ^(-1)((18^2-34^2-20^2)/(-2(34)(20))) \\ \\ C=\cos ^(-1)((-1235)/(-1360))=24.75 \end{gathered}`

VABC:

A=126.86º

B=27.07º

C=24.75º

a=34ft

b=20ft

c=18ft