*Question*

The diagram shows points in a park viewed from above, at a specific moment in time.

The distance between two trees, at points A and B, is 6.36m.

Odette is playing football in the park and is standing at point O, such that OA = 25.9m and OÂB = \(125^0\).

(a) Calculate the area of triangle AOB.

Odette’s friend, Khemil, is standing at point K such that he is 12m from A and KÂB = \(45^0\)

(b) Calculate Khemil’s distance from B.

XY is a semicircular path in the park with centre A, such that KÂY = \(45^0\). Khemil is standing on the path and Odette’s football is at point X. This is shown in the diagram below.

The length KX = 22.2m, KÔX = \(53.8^0\) and OK̂X = \(51.1^0\).

(c) Find whether Odette or Khemil is closer to the football.

Khemil runs along the semicircular path to pick up the football.

(d) Calculate the distance that Khemil runs.

**Answer/Explanation**

**Answer:**

(a) attempt to use area of triangle formula

\(\frac{1}{2} \times 25.9 \times 6.36 \times sin (125^0)\)

67.5 \(m^2\) (67.4700… \(m^2\))

(b) attempt to use cosine rule

(BK=) \(\sqrt{12^2 + 6.36^2 – 2 \times 12 \times 6.36 \times cos 45^0}\)

8.75 (m) (8.74738… (m))

(c) METHOD 1

attempt to use sine rule with measurements from triangle OKX

\(\frac{OX}{sin 51.1^o} = \frac{22.2}{sin 53.8^o}\)

(OX=) 21.4 (m) (21.4099…)(m)

(21.4 (m) < 22.2 (m))

METHOD 2

sketch of triangle OXK with vertices, angles and lengths

\(51.1^o\) is smallest angle in triangle OXK

opposite side (OX) is smallest length

therefore Odette is closest

(d) attempt to use length of arc formula

\(\frac{135}{360} \times 2 \pi \times 12 \)

28.3 (m) (9 \(\pi\), 28.2743…) (m)

*Question*

A scientist is conducting an experiment on the growth of a certain species of bacteria.

The population of the bacteria, P, can be modelled by the function

P(t) = 1200 × \(k^t\), t ≥ 0.

where t is the number of hours since the experiment began, and k is a positive constant.

(a) (i) Write down the value of P(0).

(ii) Interpret what this value means in this context.

3 hours after the experiment began, the population of the bacteria is 18 750.

(b) Find the value of k

(c) Find the population of the bacteria 1 hour and 30 minutes after the experiment began.

The scientist conducts a second experiment with a different species of bacteria.

The population of this bacteria, S, can be modelled by the function

S(t) = 5000 × \(1.65^t\)

, t ≥ 0 ,

where t is the number of hours since both experiments began.

(d) Find the value of t when the two populations of bacteria are equal.

It takes 2 hours and m minutes for the number of bacteria in the second experiment to reach 19 000.

(e) Find the value of m, giving your answer as an integer value.

**Answer/Explanation**

**Answer:**

(a) (i) 1200

(ii) the initial population of the bacteria

(b) 1200 × \(k^3\) = 18750

(k=) 2.5

(c) 1200 × \(2.5^{1.5}\)

4740 (4743.41…)

(d) equating P (t) and S (t) OR equating each function to a common variable

1200 × \(2.5^t\) = 5000 × \(1.65^t\) ; 1200 × \(2.5^t\) = x and 5000 × \(1.65^t\) = x

t = 3.43 (hours) (3.43456…)

(e) METHOD 1

5000 × \(1.65^t\) = 19000

(t=) 2.66586… OR (t – 2 ) 0.66586… (seen)

multiplying by 60 seen to convert to minutes

(m = 39.9521…)

(m=) 40 (minutes) OR 2 hours and 40 minutes

METHOD 2

equating an expression for S (t) to 19000

expressing t as 2 + \(\frac{m}{60}\)

5000 \(\times 1.65^{2 + \frac{m}{60}}\) = 19000

2 + \(\frac{m}{60}\) = 2.66586…

(m=) 40 (minutes) OR 2 hours and 40 minutes