Find the arc length of a central angle of pi/4 in a circle whose radius is 8 inches

2 answers:

7 0

The formula of the arc is:

Arc Length = 2πR.(C°/360°), where R is the radius, C°(in degrees) is the central angle of the arc:

We are given:
R = 8 in
Central angle = π/4 ≈ (180°/4) = 45°
Length of the arc= 2π x 8 x (45°/360°) →16π(1/8) :

Length of the arc = 2π in (answer B)

Solnce55 [7]2 years ago

3 0

$arc \ length =\theta*r \ \ \ [\theta= \frac{ \pi }{4}; \ \ r=8] } \\ \\ arc \ length = \frac{ \pi }{4} *8 =2 \pi \ \ in$

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Forces of 9 lbs and 13 lbs act at a 38º angle to each other. Find the magnitude of the resultant force and the angle that the re

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Answer: $R=20.84\ lb\quad 22.57^{\circ},15.43^{\circ}$

Step-by-step explanation:

Given

Two forces of 9 and 13 lbs acts $38^{\circ}$ angle to each other

The resultant of the two forces is given by

$\Rightarrow R=\sqrt{a^2+b^2+2ab\cos \theta}$

Insert the values

$\Rightarrow R=\sqrt{9^2+13^2+2(9)(13)\cos 38^{\circ}}\\\Rightarrow R=\sqrt{81+169+184.394}\\\Rightarrow R=\sqrt{434.394}\\\Rightarrow R=20.84\ lb$

Resultant makes an angle of

$\Rightarrow \alpha=\tan^{-1}\left( \dfrac{b\sin \theta}{a+b\cos \theta}\right)\\\\\text{Considering 9 lb force along the x-axis}\\\\\Rightarrow \alpha =\tan^{-1}\left( \dfrac{13\sin 38^{\circ}}{9+13\cos 38^{\circ}}\right)\\\\\Rightarrow \alpha =\tan^{-1}(\dfrac{8}{19.244})\\\\\Rightarrow \alpha=22.57^{\circ}$