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Dovator [93]
3 years ago
12

Find the exact value of csc(-1860).

Mathematics
1 answer:
Alex787 [66]3 years ago
3 0
Now, there are 360° in a circle, how many times does 360° go into 1860°?

well, let's check that,   \bf \cfrac{1860}{360}\implies \cfrac{31}{6}\implies 5\frac{1}{6}\implies 5+\frac{1}{6}

now, this is a negative angle, so it's going clockwise, like a clock moves, so it goes around the circle clockwise 5 times fully, and then it goes 1/6 extra.

well, we know 360° is in a circle, how many degrees in 1/6 of 360°?  well, is just 360/6 or their product, and that's just 60°.

so -1860, is an angle that goes clockwise, negative, 5 times fully, then goes an extra 60° passed.

5 times fully will land you back at the 0 location, if you move further down 60° clockwise, that'll land you on the IV quadrant, with an angle of -60°.

therefore, the csc(-1860°) is the same as the angle of csc(-60°), which is the same as the csc(360° - 60°) or csc(300°).

\bf csc(300^o)\implies \cfrac{1}{sin(300^o)}\implies \cfrac{1}{-\frac{\sqrt{3}}{2}}\implies -\cfrac{2}{\sqrt{3}}
\\\\\\
\textit{and if we rationalize the denominator}\qquad -\cfrac{2\sqrt{3}}{3}
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Step-by-step explanation:

See the attached graph.

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Answer:

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Step-by-step explanation:

Solution:-

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- The linear model of tree height "h" as a function of time "t" years after 2006 can be expressed in the form:

                           h = m*t + c

Where,               m = The rate of change of height

                          c = The initial height of tree in 2006.

- We will use the given data to evaluate constant "m".

                         Rate =  m = ( h2 - h1 ) / ( t2 - t1)

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-  Then use "m" and given set of data to evaluate the initial height of tree "c" in year 2006.

                          h = 2*t + c

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- The linear model for the height of tree "h" as fucntion of time "t" years after 2006 will be:

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