Question

in triangle FGH, FH = 7 ft., FG = 12 ft., and m 3 years ago

Answer 1

Is there any pictures or answer choices?

Answer 2

oh ok

Answer 3

using cosine formula
let Gh=x
\[\cos 70=\frac{ 7^2+12^2-x^2 }{ 2\times7\times12 }\]
\[168 \cos 70=49+144-x^2\]
\[x^2=193-168 \cos 70\]
\[x \approx11.64~ft\]
using cosine formula
\[\cos G=\frac{ 11.64^2+12^2-7^2 }{ 2\times12\times11.64 }\]
\[m <G=\cos^{-1} (\frac{ 11.64^2+144-49 }{ 24\times11.64})\]
G=?

Answer 4

G=34.4 degree