Question

1. a lawn sprinkler sprays water over a lawn cover an arc $160^{\circ}$ with a maximum spray distance of $x$ m as shown in the diagram below. the lawn sprinkler waters $20$ $m^{2}$ of the lawn.

(a) calculate the value of $x$. 2

1. a windscreen wiper blade is 0.8 m long. when in motion the blade moves through an arc of $\theta^{\circ}$ and wipes an area of $96$ $m^{2}$.

(a) calculate the value of $\theta$. 2

Explanation:

Step1: Use sector area formula for Q2

The area of a sector is $A = \frac{\theta}{360^\circ} \times \pi r^2$, where $\theta=160^\circ$, $A=20\ \text{m}^2$, $r=x$.
$$20 = \frac{160^\circ}{360^\circ} \times \pi x^2$$

Step2: Simplify and solve for $x$

First simplify $\frac{160}{360}=\frac{4}{9}$:
$$20 = \frac{4}{9} \pi x^2$$
Rearrange to solve for $x^2$:
$$x^2 = \frac{20 \times 9}{4\pi} = \frac{45}{\pi}$$
Take square root:
$$x = \sqrt{\frac{45}{\pi}} \approx \sqrt{14.32} \approx 3.78$$

Step3: Use sector area formula for Q3

$A=96\ \text{m}^2$, $r=0.8\ \text{m}$, solve for $\theta$:
$$96 = \frac{\theta}{360^\circ} \times \pi (0.8)^2$$

Step4: Simplify and solve for $\theta$

Calculate $(0.8)^2=0.64$:
$$96 = \frac{\theta}{360^\circ} \times 0.64\pi$$
Rearrange to isolate $\theta$:
$$\theta = \frac{96 \times 360^\circ}{0.64\pi} = \frac{34560}{2.0106} \approx 17188^\circ$$
(Note: The area value 96 m² appears disproportionately large for a 0.8 m wiper; assuming this is a typo and the area is $0.96\ \text{m}^2$, recalculating:
$$\theta = \frac{0.96 \times 360^\circ}{0.64\pi} = \frac{345.6}{2.0106} \approx 172^\circ$$)

Answer:

2 (a) $x \approx 3.78$ m
3 (a) $\theta \approx 17188^\circ$ (or $\approx 172^\circ$ if area is $0.96\ \text{m}^2$)