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Ilya [14]
3 years ago
5

Points N and R both lie on circle O. Line segment RQ is tangent to the circle at point R.

Mathematics
2 answers:
ANEK [815]3 years ago
4 0

Answer:

Perimeter ΔRON = 18.66 units

Step-by-step explanation:

Complete Question has these information given as well:

  • Circle with center O
  • RN = QN
  • RQ = 5\sqrt{3}
  • ON = 5 units
  • RQ is tangent

<em><u>Now, we draw an image according to the information given. THe image drawn is attached.</u></em>

<em><u /></em>

Looking at the figure, we can say the perimeter of triangle RON would be:

RON = OR + ON + RN

We know according to tangent theorem, QN = QR

RN is equal to that as well and RN = 5\sqrt{3}

Now,

OR and ON is the radius, which is "5"

Perimeter of RON = 5 + 5 + 5\sqrt{3} = 10 + 5\sqrt{3} = 18.66 units

<u>Perimeter ΔRON = 18.66 units</u>

LUCKY_DIMON [66]3 years ago
4 0

Answer:

18.7 units

Step-by-step explanation:

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When <em>θ</em> = <em>π</em>/3, we have <em>r</em> (<em>π</em>/3) = 13, and d<em>r</em>/d<em>θ</em> (<em>π</em>/3) = -4√3.

Differentiate these with respect to <em>θ</em> :

d<em>y</em>/d<em>θ</em> = d<em>r</em>/d<em>θ</em> sin(<em>θ</em>) + <em>r(θ)</em> cos(<em>θ</em>)

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In polar coordinates, we have

<em>y(θ)</em> = <em>r(θ)</em> sin(<em>θ</em>)

<em>x(θ)</em> = <em>r(θ)</em> cos(<em>θ</em>)

and when <em>θ</em> = <em>π</em>/3, we have <em>y</em> (<em>π</em>/3) = 13√3/2 and <em>x</em> (<em>π</em>/3) = 13/2.

The slope of the tangent line to the curve is d<em>y</em>/d<em>x</em>. By the chain rule,

d<em>y</em>/d<em>x</em> = d<em>y</em>/d<em>θ</em> • d<em>θ</em>/d<em>x</em> = (d<em>y</em>/d<em>θ</em>) / (d<em>x</em>/d<em>θ</em>)

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When <em>θ</em> = <em>π</em>/3, the slope is

d<em>y</em>/d<em>x</em> = (-4√3 sin(<em>π</em>/3) + 13 cos(<em>π</em>/3)) / (-4√3 cos(<em>π</em>/3) - 13 sin(<em>π</em>/3))

d<em>y</em>/d<em>x</em> = (-4√3 (√3/2) + 13 (1/2)) / (-4√3 (1/2) - 13 (√3/2))

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Answer:

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Step-by-step explanation:

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