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3 years ago

For a circle with the radius of 15 cm what is the length of an ark intercepted by angle measuring 120?

Mathematics

2 answers:

sp2606 [1]3 years ago

7 0

Hello from MrBillDoesMath!

Answer:

10 Pi

Discussion:

Circumference of full circle = 2 Pi r = 2 Pi (15) = 30 Pi. As a circle contains 360 degrees, 120 degrees is one third of the circle and subtends an arc of

(30 Pi)/3 = 10 Pi

Thank you,

MrB

mihalych1998 [28]3 years ago

3 0

Answer:

The length of an arc is, 31.41594 cm

Step-by-step explanation:

The arc length of the circle(l) is given by:

$l =r \theta$ ....1[]

where

r is the radius of the circle

$\theta$ is the angle in radian.

As per the statement:

For a circle with the radius of 15 cm and angle is 120 degree.

⇒r = 15 cm and angle in degree = 120

Use conversion:

1 degree = 0.0174533 radian

then;

120 degree = 2.094396 radian

⇒ $\theta = 2.094396$

Substitute the given values in [1] we have;

$l = 15 \cdot 2.094396 = 31.41594$

⇒l = 31.41594 cm

Therefore, the length of an arc is, 31.41594 cm

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$\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}$

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$\implies \displaystyle \int -e^{-2x}\sin(2x)\:\text{d}x$

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$\textsf{Let }\:u=\sin (2x) \implies \dfrac{\text{d}u}{\text{d}x}=2 \cos (2x)$

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$\begin{aligned}\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\sin (2x)- \int \dfrac{1}{2}e^{-2x} \cdot 2 \cos (2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\sin (2x)- \int e^{-2x} \cos (2x)\:\text{d}x\end{aligned}$

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Rewrite in the same format as the given integral:

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