Question

LOOK AT ATTACHMENT
MEDAL + FAN
PLEASE HELP

Answer 1

c i think

Answer 2

SOH-CAH-TOA
\[sin=\frac{opposite}{hypotenuse}\]
\[cos=\frac{adjacent}{hypotenuse}\]
\[tan=\frac{opposite}{adjacent}\]

so,\[\tan 47^{o} = \frac{a}{12}\]
Solve for a

Answer 3

none of the answers they give you look correct

Answer 4

in fact, none of them can be correct since angle A is larger than angle C...and therefore, CB must be bigger than AB...so your answer must be larger than 12

Answer 5

I got 12.9 @mtbender74

Answer 6

that's what i got too :)