Can the law of sines be used to solve the triangle
shown? Explain.
40
45°
30

Step-by-step explanation:

If BD_|_AC

in trg BDA ,sin 45=BD/40=V2/2

BD*2=40V2

BD=40V2/2=20V2  and AD=BD

so CD=30-AD=30-20V2

in trg BCD,  BC^2=BD^2+CD^2

                     BC^2=(20V2)^2+(30-20V2)^2=

                     (20V2)^2+30^2-2*30*20V2+(20V2)^2=

                     400*2+900-1200V2+400*2=

                     800+900-1200V2+800=

                      2700-1200V2

so  BC=V100(27-12V2)=10V(27-12V2

2160ft3

step-by-step explanation:

let's first notice, the center of the hyperbola is at the origin, since the focus points are equidistantly 3 units away from the origin, that also means that the distance "c" from the center to either focus is 3.

it should also be notable that, the traverse axis for this hyperbola is vertical, or namely over the y-axis and therefore, the positive fraction will be the one with y-variable.

the traverse "a" component or distance from the center to either vertex is not 2, is more than 2, so we can safely put it at √5 ≈ 2.24.