Ask question

Ask question

All categories

3 years ago

8

For a circle with a radius of 6 feet, what is the measurement of the central angle (in degrees) that intercepts an arc with a le

ngth of
9π
2
feet?

1 answer:

makkiz [27]3 years ago

3 0

Answer:

135 degrees

Step-by-step explanation:

step 1

Find the circumference of the circle

The circumference is equal to

$C=2\pi r$

we have

$r=6\ ft$

substitute

$C=2\pi (6)$

$C=12\pi\ ft$

step 2

Remember that the circumference of the circle subtends a central angle of 360 degrees

so

using proportion

Find out the measurement of the central angle (in degrees) that intercepts an arc with a length of 9π/2 ft

$\frac{12\pi}{360^o}=\frac{(9\pi/2)}{x}\\\\x=360(9\pi/2)/12\pi\\\\x=135^o$

You might be interested in

Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o

Brrunno [24]

Answer:

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

$Sin\ C = \frac{h}{a}$

Make h the subject:

$h = aSin\ C$

Also, in BOC (Using Pythagoras)

$a^2 = h^2 + x^2$

Make $x^2$ the subject

$x^2 = a^2 - h^2$

Substitute $aSin\ C$ for h

$x^2 = a^2 - h^2$ becomes

$x^2 = a^2 - (aSin\ C)^2$

$x^2 = a^2 - a^2Sin^2\ C$

Factorize

$x^2 = a^2 (1 - Sin^2\ C)$

In trigonometry:

$Cos^2C = 1-Sin^2C$

So, we have that:

$x^2 = a^2 Cos^2\ C$

Take square roots of both sides

$x= aCos\ C$

In triangle BOA, applying Pythagoras theorem, we have that:

$AB^2 = h^2 + (b-x)^2$

Open bracket

$AB^2 = h^2 + b^2-2bx+x^2$

Substitute $x= aCos\ C$ and $h = aSin\ C$ in $AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2$

Open Bracket

$AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C$

Reorder

$AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C$

Factorize:

$AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C$

In trigonometry:

$Sin^2C + Cos^2 = 1$

So, we have that:

$AB^2 = a^2 * 1 + b^2-2abCos\ C$

$AB^2 = a^2 + b^2-2abCos\ C$

Take square roots of both sides

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

6 0

3 years ago

What is the price if it’s 20% off $30

Xelga [282]

Answer: $24

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Write -2i + (9 − 3i) − (6 − 10i) as a complex number in standard form.

Vika [28.1K]

<span>-2i + (9 − 3i) − (6 − 10i)
-2i - 3i + 10i + 9 - 6
5i + 3
3 + 5i</span>

6 0

2 years ago

Read 2 more answers

Help will make you as brain

Tomtit [17]

Answer:

you still cant see it

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Tom opened a lemonade stand to earn money to buy a new bike for $200. He sells a cup of lemonade for $1.00. On the first day he

tia_tia [17]

1 - 8

2 - 16

3 - 32

x - +200

Each lemonade has a profit of $1, so the number of cups of lemonade he sells is the same as the amount of dollars he has. You just need to add each previous day's total to the next day's, doubling the amount of cups from the previous day's for each new day.

4 - (32 x 2 = ) 64

8 + 16 + 32 + 64 = 120. We need at least $80 dollars more, so we continue.

5 - (64 x 2 = ) 128

120 + 128 = $248.00 That's more than enough, so Tom will have enough money for his bike after the fifth day of selling lemonade.

6 0

2 years ago