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MatroZZZ [7]
2 years ago
13

Line BC is tangent to the circle centered at A. Find the measure of angle BCA. Explain or show your reasoning.

Mathematics
1 answer:
Virty [35]2 years ago
4 0

Answer:

m\angle BCA= 33\degree

Step-by-step explanation:

In\: \odot A BC is tangent to the circle at point B.

Therefore, by tangent-radius theorem:

\therefore AB\perp BC

\therefore \angle ABC = 90\degree

In\: \triangle ABC,

m\angle ABC +m\angle BAC +m\angle BCA= 180\degree

90\degree+57\degree +m\angle BCA= 180\degree

147\degree +m\angle BCA= 180\degree

m\angle BCA= 180\degree - 147\degree

m\angle BCA= 33\degree

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numbers of possible outcomes are the same.

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3 years ago
A cone has A height of 10.3 inches in the diameter of 6 inches what is the volume of the cone
ahrayia [7]

Answer:

V=97.08

Step-by-step explanation:

The formula for the volume of a cone is V= 1/3 pi (r)^2 (h)

To find the radius divide the diameter by 2. 6/2=3

Plugin: V=1/3 pi(3)^2(10.3)=97.08

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3 years ago
How can you write 9/20 as a decimal
Anna007 [38]
You would divide 100/20 and get 5. then u would multiply 9 times 5 equals 45 then add a decimal to get ur answer 0.45
7 0
3 years ago
Read 2 more answers
Parallelogram JKLM has vertices J(-1, 6), K(0, 9), L(6, −3), and M(3, −3). What is the coordinates of the image if the parallelo
Digiron [165]

Answer: (\dfrac13,-2),\ (0,-3),\ (-2,1),\ (-1,1) .

Step-by-step explanation:

Transformation rule for dilation:

(x,y)\to(kx,ky) , where k = scale factor

Given : Scale factor = -\dfrac13

Parallelogram JKLM has vertices J(-1, 6), K(0, 9), L(6, −3), and M(3, −3)

Vertices after dilation:

(-1,6)\to(-1\times\dfrac{-1}{3},6\times\dfrac{-1}{3})=(\dfrac13,-2)

(0,9)\times(0\times-\dfrac13,9\times-\dfrac13)=(0,-3)

(6,-3)\to(6\times-\dfrac13,\ -3\times-\dfrac13)=(-2,1)

(3,-3)\to (3\times-\dfrac13,\ -3\times-\dfrac13)=(-1,1)

Hence, the coordinates of the image if the parallelogram = (\dfrac13,-2),\ (0,-3),\ (-2,1),\ (-1,1) .

8 0
2 years ago
Need some help working these out
s2008m [1.1K]

Answers:

  • Problem 1)  40 degrees
  • Problem 2) 84 degrees
  • Problem 3) 110 degrees

===============================================

Explanation:

For these questions, we'll use the inscribed angle theorem. This says that the inscribed angle is half the measure of the arc it cuts off. An inscribed angle is one where the vertex of the angle lies on the circle, as problem 1 indicates.

For problem 1, the arc measure is 80 degrees, so half that is 40. This is the measure of the unknown inscribed angle.

Problem 2 will have us work in reverse to double the inscribed angle 42 to get 84.

-------------------

For problem 3, we need to determine angle DEP. But first, we'll need Thales Theorem which is a special case of the inscribed angle theorem. This theorem states that if you have a semicircle, then any inscribed angle will always be 90 degrees. This is a handy way to form 90 degree angles if all you have is a compass and straightedge.

This all means that angle DEF is a right angle and 90 degrees.

So,

(angle DEP) + (angle PEF) = angle DEF

(angle DEP) + (35) = 90

angle DEP = 90 - 35

angle DEP = 55

The inscribed angle DEP cuts off the arc we want to find. Using the inscribed angle theorem, we double 55 to get 110 which is the measure of minor arc FD.

3 0
3 years ago
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