Question

Find f.

A) 7.4

B) 8.2

C) 10.5

D) 11.1

Answer 1

F=72

g=6

------------


`\cos { \left( F \right) } =\frac { { e }^( 2 )+{ g }^( 2 )-{ f }^( 2 ) }{ 2eg }`

Therefore:


`\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 }`


`\\ \\ 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 }`

But what is e?

E=76

G=32

g=6

And:


`\frac { e }{ \sin { \left( E \right) } } =\frac { g }{ \sin { \left( G \right) } }`

Which means that:


`\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) } }`

If you take this value into account, you will discover that f is...


`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)`

So I would have to say that the answer is approximately (c).