Answer:
see explanation
Step-by-step explanation:
Using the tangent and sine ratios in the right triangle EFG
tan60° = 
= 
= 
( multiply both sides by EG )
EG × tan60° = 28 ( divide both sides by tan60° )
EG = 
≈ 16.2 in ( to the nearest tenth )
--------------------------------------------------------------
sin60° = 
= 
= 
( multiply both sides by EF )
EF × sin60° = 28 ( divide both sides by sin60° )
EF = 
≈ 32.3 in ( to the nearest tenth )
I’m bad at math too sorry i can’t helppp i just need to answer one to ask a question
Using the equations for tangent and secant lines:
Let the unknown length of the line inside the circle = y
6^2 = 3 x (y+3)
Simplify:
36 = 3y+9
Subtract 9 from both sides:
27 =3y
Divide both sides by 3:
Y = 9
Now add to get x:
X = 9 + 3 = 12
X = 12
You are looking for the shaded region that would be contained in both of the inequalities.
You have:
If you graph an shade the correct half-plane for those equations, you will see there is a triangular region on the left side of the first quadrant.
