F=72
g=6
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[tex]\cos { \left( F \right) } =\frac { { e }^{ 2 }+{ g }^{ 2 }-{ f }^{ 2 } }{ 2eg } [/tex]
Therefore:
[tex]\cos { \left( 72 \right) } =\frac { { e }^{ 2 }+{ 6 }^{ 2 }-{ f }^{ 2 } }{ 2\cdot e\cdot 6 } \\ \\ \cos { \left( 72 \right) } =\frac { { e }^{ 2 }+36-{ f }^{ 2 } }{ 12e }[/tex]
[tex]\\ \\ 12e\cdot \cos { \left( 72 \right) } ={ e }^{ 2 }+36-{ f }^{ 2 }\\ \\ \therefore \quad { f }^{ 2 }={ e }^{ 2 }-12e\cdot \cos { \left( 72 \right) } +36\\ \\ \therefore \quad f=\sqrt { { e }^{ 2 }-12e\cdot \cos { \left( 72 \right) +36 } } \\ \\ \therefore \quad f=\sqrt { e\left( e-12\cos { \left( 72 \right) } \right) +36 } [/tex]
But what is e?
E=76
G=32
g=6
And:
[tex]\frac { e }{ \sin { \left( E \right) } } =\frac { g }{ \sin { \left( G \right) } } [/tex]
Which means that:
[tex]\frac { e }{ \sin { \left( 76 \right) } } =\frac { 6 }{ \sin { \left( 32 \right) } } \\ \\ \therefore \quad e=\frac { 6\cdot \sin { \left( 76 \right) } }{ \sin { \left( 32 \right) } } [/tex]
If you take this value into account, you will discover that f is...
[tex]f=\sqrt { \frac { 6\cdot \sin { \left( 76 \right) } }{ \sin { \left( 32 \right) } } \left( \frac { 6\cdot \sin { \left( 76 \right) } }{ \sin { \left( 32 \right) } } -12\cos { \left( 72 \right) } \right) +36 } \\ \\ \therefore \quad f=10.8\quad \left( 1\quad d.p \right) [/tex]
So I would have to say that the answer is approximately (c).
