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vampirchik [111]
3 years ago
7

A circular plate has a crack along the line AB, as shown below: A circular plate is shown. The center of the plate is labeled as

A. B is a point on the circumference of the plate. A straight line joins the points A and B. If A is the center of the plate, what is the segment AB called?
Mathematics
2 answers:
makkiz [27]3 years ago
8 0
Well, since point A represents the center, and B represents a point on the outer line of the circle, segment AB would represent the radius, since the radius represents the length from the center to the outside of a circle

Hope this helps
deff fn [24]3 years ago
6 0

Answer:

Segment AB is called the Radius.

Step-by-step explanation:

We are given a circular plate which has a crack along the line AB.

The center of the plate is labeled as A and B is a point on the circumference of the plate.

Now we are asked to find what the line segment AB is called.

We know that radius of a circle is a line segment that joins the center of the circle to any point on the circumference of the circle.

Here we have Point A is the center and B is the point on the circumference.

Hence, Line AB is called the Radius of the circle.

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3/2

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sin(3π/4 - β) = sin(3π/4)cosβ - cos(3π/4)sinπ =

\sin(\frac{3\pi}{4}-\beta)=\sin(\frac{3\pi}{4})\cos\beta-\cos\frac{3\pi}{4}\sin\beta\\=\frac{1}{\sqrt{2}}\cos\beta-(-\frac{1}{\sqrt{2}})\sin\beta\\\\=\frac{1}{\sqrt{2}}(\cos\beta +\sin\beta)\\\\\sin^2(\frac{3\pi}{4}-\beta)=\frac{1}{2}\cos\beta +\sin\beta)^2=\frac{1}{2}(\cos^2\beta +\sin^2\beta+2\sin\beta\cos\beta\\=\frac{1}{2}(1+\sin2\beta)=\frac{1}{2}(1-\frac{1}{5}) = \frac{2}{5}\\

Use \cot^2x=\csc^2x-1=\frac{1}{\sin^2x}-1

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