Question

Abcd is a parellogram and AE and CF bisect angle A and Angle C. prove that AE||FC.​

Answer 1

$\huge\mathcal\blue{Answer}$

14.

∫ A sin (ωt) dt = (A/ω) ∫ sin (ωt) d(ωt)

∫ A sin (ωt) dt = - (A/ω) cos (ωt)

Applying limit for t from 0 to t,

∫ A sin (ωt) dt = - (A/ω) [cos (ωt) - cos (ω*0)]

∫ A sin (ωt) dt = - (A/ω) [cos (ωt) - 1]

15.

v dv = - ω² x dx

∫ v dv = - ω² ∫ x dx

Integral is done for velocity from v0 to v.

Integral is done for displacement from 0 to x.

Therefore,

[v² - (v0)²] / 2 = - ω² [x² - 0²] / 2

v² = (v0)² - ω²x²

16.

dq = e^(-t / τ) dt

q = ∫ e^(-t / τ) dt

q = - τ ∫ e^(-t / τ) d(-t / τ)

q = - τ e^(-t / τ)

Applying limit for t from 0 to τ,

q = - τ [e^(-τ / τ) - e^(-0 / τ)]

q = τ (1 - 1/e)

Answer 2

Answer:

∫ A sin (ωt) dt = (A/ω) ∫ sin (ωt) d(ωt)

∫ A sin (ωt) dt = - (A/ω) cos (ωt)

Applying limit for t from 0 to t,

∫ A sin (ωt) dt = - (A/ω) [cos (ωt) - cos (ω*0)]

∫ A sin (ωt) dt = - (A/ω) [cos (ωt) - 1]

15.

v dv = - ω² x dx

∫ v dv = - ω² ∫ x dx

Integral is done for velocity from v0 to v.

Integral is done for displacement from 0 to x.

Therefore,

[v² - (v0)²] / 2 = - ω² [x² - 0²] / 2

v² = (v0)² - ω²x²

16.

dq = e^(-t / τ) dt

q = ∫ e^(-t / τ) dt

q = - τ ∫ e^(-t / τ) d(-t / τ)

q = - τ e^(-t / τ)

Applying limit for t from 0 to τ,

q = - τ [e^(-τ / τ) - e^(-0 / τ)]

q = τ (1 - 1/e)