Question 1. [Maximum mark: 17]

The diagram below shows a circular clock face with centre O. The clock’s minute hand has a length of 10 cm. The clock’s hour hand has a length of 6 cm.

At 4:00 pm the endpoint of the minute hand is at point A and the endpoint of the hour hand is at point B.

a. Find the size of angle AÔB in degrees. 

b.  Find the distance between points A and B. 

Between 4:00 pm and 4:13 pm, the endpoint of the minute hand rotates through an angle, θ ,

from point A to point C. This is illustrated in the diagram.

                     

c.  Find the size of angle θ in degrees. 

d.  Calculate the length of arc AC. 

e.  Calculate the area of the shaded sector, AOC. 

A second clock is illustrated in the diagram below. The clock face has radius 10 cm with minute and hour hands both of length 10 cm. The time shown is 6:00 am. The bottom of the clock face is located 3 cm above a horizontal bookshelf.

                               

Write down the height of the endpoint of the minute hand above the bookshelf at 6:00 am. 

f.    The height, h centimetres, of the endpoint of the minute hand above the bookshelf is modelled by the function

h(θ ) = 10 cos θ + 13 , θ ≥ 0 ,

where θ is the angle rotated by the minute hand from 6:00 am.

g.   Find the value of h when θ = 160° . 

The height, g centimetres, of the endpoint of the hour hand above the bookshelf is modelled by the function

  g(θ ) = -10 cos (θ/12) + 13 , θ ≥ 0 ,

where θ is the angle in degrees rotated by the minute hand from 6:00 am.

h.   Write down the amplitude of g(θ ) . 

The endpoints of the minute hand and hour hand meet when θ = k .

Find the smallest possible value of k . 

Answer/Explanation

(a) \(4\times \frac{360°\cdot }{12}OR 4\times 30° = 120°\)

(b)  cosine rule \(AB^{2}= 10^{2}+6^{2}-2\times 10\times 6\times cos (120°)\)

\(AB= 14cm\)

(c) \(\Theta = 13\times 6= 78°\)

(d) arc length l= \(\frac{78}{360} \times 2\times \pi \times 10\) OR

\(l= \frac{13\pi }{30}\times 10 = 13.6cm(13.6135…,4.33\pi ,\frac{13\pi }{3})\)

(e) Area of a sector \(A= \frac{78}{360}\times \pi \times 10^{2}\) OR \( l=\frac{1}{2}\times \frac{13\pi }{30}\times 10^{2}= 68.1cm^{2}(68.0678…,21.7\pi ,\frac{65\pi }{3}\)

(f) 23

(g) correct substitution h =10cos (160 ) + 13 = 3.60 cm (3.60307…)

(h) 10 (i) EITHER 10\(\times cos(\Theta )+13 = -10\times cos(\frac{\Theta }{12})\)+13

OR THEN k =196° (196.363…)