Find the angle of arc BC

Mathematics

1 answer:

amid [387]3 years ago

6 0

Answer:

96°

Step-by-step explanation:

The measure of arc EC is twice the measure of ∠DEC, so is ...

arc EC = 2×68° = 136°

The measure of arc BC is found from the sum of arcs around a circle.

BC +CE +EB = 360° = BC +136° +128°

BC = 96° . . . . . subtract 264°

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Simplifying the integrand, we have

$\vec F\cdot d\vec r = \bigg(-\cos(t)\sin(t) + \sin(t)\cos(t) + \sin(t)\cos(t)\bigg) \, dt \\ ~~~~~~~~= \sin(t)\cos(t) \, dt \\\\ ~~~~~~~~= \dfrac12 \sin(2t) \, dt$

Then the line integral evaluates to

$\displaystyle \int_C \vec F\cdot d\vec r = \int_0^\pi \frac12\sin(2t)\,dt \\\\ ~~~~~~~~ = -\frac14\cos(2t) \bigg|_{t=0}^{t=\pi} \\\\ ~~~~~~~~ = -\frac14(\cos(2\pi)-\cos(0)) = \boxed{0}$