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ankoles [38]
3 years ago
15

Need to get the values of each trig function using half angle formulas within the constraint

Mathematics
1 answer:
Marizza181 [45]3 years ago
8 0
\bf cot(u)=4\qquad \pi \ \textless \ u\ \textless \ \frac{3\pi }{2}\implies cot=\cfrac{-4}{-1}\cfrac{\leftarrow adjacent}{\leftarrow opposite}
\\\\\\
hypotenuse\implies \sqrt{(-4)^2+(-1)^2}\implies \sqrt{17}
\\\\\\
cos(\theta )=\cfrac{-4}{\sqrt{17}}\implies \cfrac{-4\sqrt{17}}{17}\\\\
-------------------------------\\\\


\bf sin\left( \frac{u}{2} \right)=\pm\sqrt{\cfrac{1+\frac{4\sqrt{17}}{17}}{2}}\implies \pm\sqrt{\cfrac{\frac{17+4\sqrt{17}}{17}}{2}}\implies \boxed{\pm\sqrt{\cfrac{17+4\sqrt{17}}{34}}}
\\\\\\
cos\left( \frac{u}{2} \right)=\pm\sqrt{\cfrac{1-\frac{4\sqrt{17}}{17}}{2}}\implies \pm\sqrt{\cfrac{\frac{17-4\sqrt{17}}{17}}{2}}\implies \boxed{\pm\sqrt{\cfrac{17-4\sqrt{17}}{34}}}

\bf tan\left( \frac{u}{2} \right)=\pm\sqrt{\cfrac{\frac{17+4\sqrt{17}}{34}}{\frac{17-4\sqrt{17}}{34}}}\implies \boxed{\pm\sqrt{\cfrac{17+4\sqrt{17}}{17-4\sqrt{17}}}}
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Problem 1

x = measure of angle N

2x = measure of angle M, twice as large as N

3(2x) = 6x = measure of angle O, three times as large as M

The three angles add to 180 which is true of any triangle.

M+N+O = 180

x+2x+6x = 180

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x = 180/9

x = 20 is the measure of angle N

Use this x value to find that 2x = 2*20 = 40 and 6x = 6*20 = 120 to represent the measures of angles M and O in that order.

<h3>Answers:</h3>
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  • Angle N = 20 degrees
  • Angle O =  120 degrees

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

Problem 2

n = number of sides

S = sum of the interior angles of a polygon with n sides

S = 180(n-2)

2700 = 180(n-2)

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n = 15+2

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Problem 3

x = smaller acute angle

3x = larger acute angle, three times as large

For any right triangle, the two acute angles always add to 90.

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x = 22.5

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<h3>Answers:</h3>
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