Please help with a trigonometry question.?

Question

A circle of radius r is divided into n equal segments to form pie-shaped segment ABC where the angle A at the center of the circle is 360/n and the length of arc BC is  2 π r/n.

Now extend legs AB and AC beyond distance r. Place line DE tangent to arc BC to form an equilateral triangle ADE. How do I find the length of line DE?

My only Trigonometry course was over 30 years ago and I have not used the concepts since then. Thanks for the help.

Puzzling · Answer

As described, you have an equilateral triangle with a height equal to the radius (r) of the circle. You can divide that into two right triangles. Using the ratios of a 30-60-90 triangle:
short leg : hypotenuse : long leg
1 : 2 : √3
Multiplying those all by r and dividing by √3 you get
r/√3 : 2r/√3 : r

In other words for a height (long leg) equal to the radius (r), the side of the equilateral triangle is the same as 2r/√3 (the hypotenuse).

We can rationalize that value by multiplying by √3/√3 so we don't have a square root in the denominator.
= (2√3)r/3

Answer: 
DE is (2√3)r/3
(approximately  1.1547 times the radius)

? · Answer

Bisect angle A to point of DE tangency.

Two right triangles will form.

o = 1/2 DE
a = r
theta = x = (0.5) 2pi/n = pi/n

tan (x) = o/a
o = a tan (x)
o = r * tan (pi/n)
DE = 2 * o
DE = 2r tan (pi/n)

The way I read your question ADE is definitely an isosceles triangle, not necessarily equilateral.
Sorry, if I have misunderstood and the answer above is not valid.

Myles · Answer

Triangle ADE has height equal to the radius of the circle. You also know the angle at the centre. That's all you need.

? · Answer

Actually, that pie-shaped piece is a sector, not a segment.

∠ADE = (180 - 360/n)/2 = 180/2 - 360/(2n) = 90 - 180/n

tan(90-180/n) = r/(DE/2) = 2r/DE
DE = 2r/tan(90-180/n)

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Please help with a trigonometry question.?

4 Answers