A disc is cut into 12 sectors with areas in arithmetic progression. The largest angle is twice the smallest. Find the smallest angle.

Question

I was given a question which states -

A circular disc is cut into twelve sectors whose areas are in an arithmetic sequence. The angle of the largest sector is twice the angle of the smallest sector.

Find the size of the angle of the smallest sector.

Here's what I have done so far,

$\Rightarrow U_1 = w$

$\Rightarrow U_n = 2w$

$\Rightarrow w + (n - 1)d = 2w$

$\Rightarrow n = 12$

$\therefore\ d = \frac{w}{11}$

And now I am clueless about where to go from here...

Answer 1

Recall that the sum of twelve sectors is the whole disc, $$360^{\circ}=\sum_{k=1}^{12} (w+(k-1)d)=12w+d\sum_{k=1}^{12}(k-1)=12w+66d$$ which, together with $w + (12 - 1)d = 2w$, i.e. $11d=w$ give you all you need to find $w$, that is the angle (in degrees) of the smallest sector: $$360^{\circ}=12w+66d=12w+6w\implies w=\frac{360^\circ}{18}=20^\circ.$$

Answer 2

You got $d = a$₁$/11$.

($a$₁ is angle of the smallest sector)

We can write: $11d = a$₁ ........(1)

Now, we know:

Area of sector = ${\frac{1}2}.{r^2}.\theta$

And,

Sum of area of all sectors = Area of circle

(${\frac{1}2}.{r^2}.a$₁) + (${\frac{1}2}.{r^2}.a$₂) + ..... + (${\frac{1}2}.{r^2}.a$₁₂) = $\pi.r^2$

${\frac{1}2}.r^2$.{ $a$₁ + $a$₂ + ..... + $a$₁₂ } = $\pi.r^2$

$a$₁ + $a$₂ + ..... + $a$₁₂ = 2$\pi$

Applying the formula for the sum of $n$ terms of an AP:

${\frac{12}2}.${ $2a$₁ + $(12-1).d$ } = 2$\pi$

$6$.{ $2a$₁ + $11d$ } = 2$\pi$

Putting value of $11d$ from (1):

$6$.{ $2a$₁ + $a$₁ } = 2$\pi$

$a$₁ = $\frac{\pi}9$

Answer 3

You could also use the formula -

$\frac{n}{2}(U_1 + U_n)$

Like this,

$\Rightarrow\ \frac{12}{2}(w^{\circ} + 2w^{\circ}) = 360^{\circ}$

$\Rightarrow\ 6(3w^{\circ}) = 360^{\circ}$

$\Rightarrow\ w^{\circ} = \frac{360^{\circ}}{18}$

$\therefore\ w^{\circ} = 20^{\circ}$