Is the given point interior, exterior, or on the circle (x + 2)2 + (y - 3)2 = 81? P(8, 4)
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Given that ∠B ≅ ∠C.
to prove that the sides AB = AC
This can be done by the method of contradiction.
If possible let AB =AC
Then either AB>AC or AB<AC
Case i: If AB>AC, then by triangle axiom, Angle C > angle B.
But since angle C = angle B, we get AB cannot be greater than AC
Case ii: If AB<AC, then by triangle axiom, Angle C < angle B.
But since angle C = angle B, we get AB cannot be less than AC
Conclusion:
Since AB cannot be greater than AC nor less than AC, we have only one possibility. that is AB =AC
Hence if angle B = angle C it follows that
AB = AC, and AB ≅ AC.
The attached figure represents the relation between ω (rpm) and t (seconds)
To find the blade's angular position in radians ⇒ ω will be converted from (rpm) to (rad/s)
ω = 250 (rpm) = 250 * (2π/60) = (25/3)π rad/s
ω = 100 (rpm) = 100 * (2π/60) = (10/3)π rad/s
and from the figure it is clear that the operation is at constant speed but with variable levels
⇒ ω = dθ/dt ⇒ dθ = ω dt
∴ θ = ∫₀²⁰ ω dt
while ω is not fixed from (t = 0) to (t =20)
the integral will divided to 3 integrals as follow;
ω = 0 from t = 0 to t = 5
ω = 250 (rpm) = (25/3)π from t = 5 to t = 15
ω = 100 (rpm) = (10/3)π from t = 15 to t = 20
∴ θ = ∫₀⁵ (0) dt + ∫₅¹⁵ (25/3)π dt + ∫₁₅²⁰ (10/3)π dt
the first integral = 0
the second integral = (25/3)π t = (25/3)π (15-5) = (250/3)π
the third integral = (10/3)π t = (25/3)π (20-15) = (50/3)π
∴ θ = 0 + (250/3)π + (50/3)π = 100 π
while the complete revolution = 2π
so instantaneously at t = 20
∴ θ = 100 π - 50 * 2 π = 0 rad
Which mean:
the blade will be at zero position making no of revolution = (100π)/(2π) = 50
To find the blade's angular position in radians ⇒ ω will be converted from (rpm) to (rad/s)
ω = 250 (rpm) = 250 * (2π/60) = (25/3)π rad/s
ω = 100 (rpm) = 100 * (2π/60) = (10/3)π rad/s
and from the figure it is clear that the operation is at constant speed but with variable levels
⇒ ω = dθ/dt ⇒ dθ = ω dt
∴ θ = ∫₀²⁰ ω dt
while ω is not fixed from (t = 0) to (t =20)
the integral will divided to 3 integrals as follow;
ω = 0 from t = 0 to t = 5
ω = 250 (rpm) = (25/3)π from t = 5 to t = 15
ω = 100 (rpm) = (10/3)π from t = 15 to t = 20
∴ θ = ∫₀⁵ (0) dt + ∫₅¹⁵ (25/3)π dt + ∫₁₅²⁰ (10/3)π dt
the first integral = 0
the second integral = (25/3)π t = (25/3)π (15-5) = (250/3)π
the third integral = (10/3)π t = (25/3)π (20-15) = (50/3)π
∴ θ = 0 + (250/3)π + (50/3)π = 100 π
while the complete revolution = 2π
so instantaneously at t = 20
∴ θ = 100 π - 50 * 2 π = 0 rad
Which mean:
the blade will be at zero position making no of revolution = (100π)/(2π) = 50