Need help trigonometry

The measure of angle F is 67.6°

Step-by-step explanation:

In the right triangle DEF

∵ ∠E is the right angle

∵ Side DF is opposite to ∠E

∴ DF is the hypotenuse of the triangle DEF

∵ Side EF is the adjacent side of ∠F

∵ The cosine ratio =

∴ cos∠F =

∵ EF = 8 units

∵ DF = 21 units

→ Substitute them in the cosine ratio above

∴ cos∠F =

→ Use to find the measure of angle F

∵ m∠F =

∴ m∠F = 67.60731219 degrees

→ Round it the nearest tenth of a degree

∴ m∠F = 67.6°

∴ The measure of angle F is 67.6°

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)