Abcd is a cyclic quadrilateral. o is the center of the circle ∠bad = 40°,then angle BCD=?,m(arc BCD) =?,m( arc BAD)=?

m∠BCD = 140°

m(arc BCD) = 80°

m(arc BAD) = 280°

Step-by-step explanation:

Let us use these facts to solve the question

∵ ABCD is a cyclic quadrilateral in circle O

∵ ∠BAD and ∠BCD are opposite angles

→ By using the 1st fact above

∴ m∠BAD + m∠BCD = 180°

∵ m∠BAD = 40°

∴ 40° + m∠BCD = 180°

→ Subtract 40° from both sides

∴ 40° - 40° + m∠BCD = 180° - 40°

∴ m∠BCD = 140°

∵ Arc BCD subtended the inscribed ∠BAD

→ By using the 2nd fact above

∴ m(arc BCD) = 2(m∠BAD)

∵ m∠BAD = 40°

∴ m(arc BCD) = 2(40°)

∴ m(arc BCD) = 80°

∵ Arc BAD subtended the inscribed ∠BCD

→ By using the 2nd fact above

∴ m(arc BAD) = 2(m∠BCD)

∵ m∠BCD = 140°

∴ m(arc BAD) = 2(140°)

∴ m(arc BAD) = 280°.

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step-by-step explanation:

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