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Question Number 6758 by FilupSmith last updated on 23/Jul/16

A circle of radius r has a point O as its

centre . Points A and B are points on the

circumference .

For △ O A B, \overset{―}{O A} = \overset{―}{O B} = r, \overset{―}{A B} = d, ∠ A O B = θ .

What is \frac{r}{d} ?

Commented by FilupSmith last updated on 23/Jul/16

I^{'} ve used the cosine rule and pythagorus

and have gotten two different results for

the variable d .

Commented by Rasheed Soomro last updated on 23/Jul/16

B y c o s i n e l a w

d^{2} = r^{2} + r^{2} - 2 r^{2} c o θ = 2 r^{2} (1 - c o s θ)

\frac{r^{2}}{d^{2}} = \frac{1}{2 (1 - c o s θ)}

\frac{r}{d} = + \sqrt{\frac{1}{2 (1 - c o s θ)}} [S i n c e r, d > 0, \frac{r}{d} > 0]

= \frac{1}{\sqrt{2 (1 - c o s θ)}} \dots \dots \dots \dots \dots \dots \dots \dots A

- - - - - - - - -

B y P y t h a g o r u s t h e o r m

s i n \frac{θ}{2} = \frac{d / 2}{r} = \frac{d}{2 r}

\frac{d}{r} = 2 s i n \frac{θ}{2}

B u t s i n \frac{θ}{2} = \pm \sqrt{\frac{1 - c o s θ}{2}}

H e n c e \frac{d}{r} = 2 (+ \sqrt{\frac{1 - c o s θ}{2}}) [S i n c e r, d > 0, \frac{r}{d} > 0]

\frac{r}{d} = \frac{1}{2} (\frac{\sqrt{2}}{\sqrt{1 - c o s θ}}) = \frac{1}{\sqrt{2}} (\frac{1}{\sqrt{1 - c o s θ}})

= \frac{1}{\sqrt{2 (1 - c o s θ)}} \dots \dots \dots \dots \dots \dots \dots . . B

B o t h r e s u l t s a r e s a m e .

Commented by FilupSmith last updated on 24/Jul/16

Thank you! I failed to realise that

\frac{d}{r} = 2 \sin (\frac{θ}{2})

Commented by sandy_suhendra last updated on 25/Jul/16

\frac{r}{d} = \frac{1}{\sqrt{2 (1 - c o s θ)}}

\frac{d}{r} = \sqrt{2 (1 - c o s θ)} \Rightarrow 1 - c o s θ = 2 {s i n}^{2} (\frac{θ}{2})

\frac{d}{r} = \sqrt{4 {s i n}^{2} (\frac{θ}{2})} = 2 s i n (\frac{θ}{2})