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max2010maxim [7]

3 years ago

13

A circle has a circumference of 12 it has an arc of length 11 what is the central angle of the arc in degreesA circle has a circ

umference of 12 it has an arc of length 11 what is the central angle of the arc in degrees

2 answers:

Ilia_Sergeevich [38]3 years ago

7 0

Answer:

330°

Step-by-step explanation:

The central angle:

360°× $\frac{11}{12}$

=30°×11

=330°

IrinaVladis [17]3 years ago

4 0

Answer: 330⁰

<u>Step-by-step explanation:</u>

Circumference (C) = 2π · radius(r)

12 = 2π·r

$\dfrac{6}{\pi}=r$

Arc length (s) = radius(r) · θ <em>(reminder that theta is in radians)</em>

$11=\dfrac{6}{\pi}\cdot \theta\\$

$\dfrac{11\pi}{6}=\theta$

To convert radians to degrees, use the proportion: π = 180⁰

$\dfrac{\pi}{180}=\dfrac{11\pi}{6x}\\\\6x(\pi)=180(11\pi)\\\\x=\dfrac{180(11\pi)}{6\pi}\\\\x=30(11)\\\\x=330$

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4 0

3 years ago

Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

Ludmilka [50]

Answer:

$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

$(c)\ \cot(\frac{\pi}{2} - \theta) = 9$

$(d)\ \csc^2(\theta) = \frac{82}{81}$

Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

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<h3>#1</h3>

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A = lw

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<h3>#2</h3>

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P = 2 1/4 + 5 2/5 + 5 17/20 =

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<h3>#3</h3>

Cost of a shirt = $19.99
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<u>Total cost:</u>

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