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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: bubbles on May 12, 2008, 09:34:26 pm

Title: Circular functions problem (Essentials)
Post by: bubbles on May 12, 2008, 09:34:26 pm: Need help with these two questions:

Q14 pg 127
A curve on a light rail track is an arc of a circle of length 300m and the straight line joining the two ends of the curve is 270m long.
a) show that, if the arc subtends an angle of 2θ° at the centre of the circle, θ is a solution of the equation
sinθ° = [π / (200)]θ°
b) solve, correct to two decimal places, the equation for θ.

Q19 pg 128
A string is wound around a disc and a horizontal length of the string AB is 20cm long. The radius of the disc is 10cm. The string is then moved so that the end of the string B', is moved to a point at the same level as O, the centre of the circle. B'P is a tangent to the circle.
a) show that θ satisfies the equation (π/2) - θ + tan θ = 2
b) Find the value of θ, correct to two decimal places, which satisfies this equation.
Title: Re: Circular functions problem (Essentials)
Post by: Mao on May 12, 2008, 10:33:01 pm: Q14:
Quote from: Ahmad on February 11, 2008, 05:00:39 pm
Ratio of angles = Ratio of arc length to circumference ( $\theta$ in degrees)
$\implies \frac{2 \theta}{360} = \frac{300}{2\pi r}$

$\implies \sin \theta = \frac{270/2}{r} = \frac{\pi}{200} \theta$
Title: Re: Circular functions problem (Essentials)
Post by: bubbles on May 12, 2008, 10:44:24 pm: Thanks Mao!
Title: Re: Circular functions problem (Essentials)
Post by: Mao on May 12, 2008, 11:00:44 pm: Looking back, i dont actually understand how that worked again clearly, so i feel obliged to post a more detailed solution:

14:
imagine the sector with the angle $2\theta$ , arc length of 300, and a straight line across it of 270

cutting that down the middle will give a right-angled-triangle, with $\theta$ as the angle, 135 as its opposite, and r as its hypotenuse, that is:

$\sin \theta = \frac{135}{r}$

now, from our sector, as Ahmad has pointed out 3 months + 1 day ago, that:

Ratio of angles = Ratio of arc length to circumference ( $\theta$ in degrees)
$\implies \frac{2 \theta}{360} = \frac{300}{2\pi r}$

$\implies r = \frac{300 \cdot 360}{4 \cdot \pi \cdot \theta}$

substituting this:

$\sin \theta = \dfrac{135}{\left(\dfrac{300 \cdot 360}{4 \cdot \pi \cdot \theta}\right)} =\frac{135 \cdot 4 \cdot \pi \cdot \theta}{300 \cdot 360} =\frac{\pi}{200}\cdot \theta$

Q19
this question is worded really ambiguously [and drawn ambiguously as well!]

what it bascally asks is:
(http://obsolete-chaos.wikispaces.com/space/showimage/arc.JPG)

so:

$Arc_{AP} + PB' = 20$ , firstly, we realise that the Arc length is angle * radius

$\implies \left( \frac{\pi}{2} - \theta \right) \cdot r + PB' = 20$ , then, we realise that $\tan(\theta) = \frac{PB'}{r} \implies PB' = \tan(\theta) \cdot r$

$\implies \left( \frac{\pi}{2}-\theta \right) \cdot 10 + \tan(\theta) \cdot 10 = 20$

$\implies \frac{\pi}{2}-\theta + tan(\theta) = 2$