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SVEN [57.7K]
3 years ago
11

How would I go about solving this?

Mathematics
1 answer:
garik1379 [7]3 years ago
8 0
If you draw more of those triangles, there will be 6 that can fit, so find area of 1 triangle and multiply it by 6. Write that number down and then do Pi r squared to find the area of the circle, then do circle area minus triangles area when you get that, divide it by 6. That is the area of the white region so then do Pi R squared again and then subtract the white area from that
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Step-by-step explanation:

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tigry1 [53]

Step-by-step explanation:

\frac{( \csc x+1)( \csc  x-1)}{ \csc ^2x} =  \frac{ { \csc }^{2} x - 1}{  \csc^{2} x}  \\ = 1 -   \sin^{2} x =   \cos^{2} x \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

b. Since csc theta = 1/sin theta, we can multiply both sides by sin theta and you will end up with

\cos^{2} \theta  +  \sin^{2} \theta  = 1

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3 years ago
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Lubov Fominskaja [6]

Answer:

See Below.

Step-by-step explanation:

We are given the isosceles triangle ΔABC. By the definition of isosceles triangles, this means that ∠ABC = ∠ACB.

Segments BO and CO bisects ∠ABC and ∠ACB.

And we want to prove that ΔBOC is an isosceles triangle.

Since BO and CO are the angle bisectors of ∠ABC and ∠ACB, respectively, it means that ∠ABO = ∠CBO and ∠ACO = ∠BCO.

And since ∠ABC = ∠ACB, this implies that:

∠ABO = ∠CBO =∠ACO = ∠BCO.

This is shown in the figure as each angle having only one tick mark, meaning that they are congruent.

So, we know that:

\angle ABC=\angle ACB

∠ABC is the sum of the angles ∠ABO and ∠CBO. Likewise, ∠ACB is the sum of the angles ∠ACO and ∠BCO. Hence:

\angle ABO+\angle CBO =\angle ACO+\angle BCO

Since ∠ABO =∠ACO, by substitution:

\angle ABO+\angle CBO =\angle ABO+\angle BCO

Subtracting ∠ABO from both sides produces:

\angle CBO=\angle BCO

So, we've proven that the two angles are congruent, thereby proving that ΔBOC is indeed an isosceles triangle.

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