Question 1 · Circular sector: the sprinkler
Degrees to radians, arc length and area of an annular sector; angle for a given area
Maximum mark · 12 marksA garden sprinkler is fixed at a point $O$ in a field. The jet reaches a maximum distance of $9$ m and the arm rotates through an angle of $150^\circ$. Because of the way it is built, the sprinkler does not water the region within $2$ m of $O$.
- Write the angle of rotation in radians, giving the exact answer in terms of $\pi$. 2 p
- Find the length of the arc traced by the end of the jet. 3 p
- Find the area of the watered region. 4 p
- The sprinkler is to be adjusted so that it waters an area of $150$ m², keeping the $9$ m reach and the $2$ m dry zone. Find the new angle of rotation, in radians and in degrees. 3 p
IB AI HL
TANS 3.7
Step-by-step solution
a) Degrees to radians
$$150^\circ = 150\cdot\dfrac{\pi}{180} = \dfrac{5\pi}{6}\ \text{rad}$$
Conversion factor 1 mark, exact answer 1 mark.
b) Arc length
$$\ell = r\theta = 9\cdot\dfrac{5\pi}{6} = \dfrac{15\pi}{2} \approx 23.6\ \text{m}$$
Formula 1 mark, substitution 1 mark, answer 1 mark.
c) Watered area
It is the difference between two sectors with the same angle:
$$A = \dfrac{1}{2}\left(R^2 - r^2\right)\theta = \dfrac{1}{2}\left(9^2 - 2^2\right)\cdot\dfrac{5\pi}{6} = \dfrac{385\pi}{12} \approx 101\ \text{m}^2$$
Idea of subtracting the two sectors 2 marks, substitution 1 mark, answer 1 mark.
d) New angle
$$\dfrac{1}{2}\,(81-4)\,\theta = 150 \;\Longrightarrow\; \theta = \dfrac{300}{77} \approx 3.90\ \text{rad} \approx 223^\circ$$
$\theta \approx 3.90$ rad $\approx 223^\circ$
Setting up 1 mark, radians 1 mark, degrees 1 mark.