Home / IBDP Maths SL 3.5 Definition of cos , sin and tan angles AA HL Paper 2- Exam Style Questions

IBDP Maths SL 3.5 Definition of cos , sin and tan angles AA HL Paper 2- Exam Style Questions

IBDP Maths SL 3.5 Definition of cos , sin and tan angles AA HL Paper 2- Exam Style Questions- New Syllabus

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Question

A rotating sprinkler is at a fixed point \( S \). It waters all points inside and on a circle of radius 20 meters. Point \( S \) is 14 meters from the edge of a path which runs in a north-south direction. The edge of the path intersects the circle at points \( A \) and \( B \).

Sprinkler Diagram 1

a. Show that \( AB = 28.57 \), correct to four significant figures. [3]

b. The sprinkler rotates at a constant rate of one revolution every 16 seconds. Show that the sprinkler rotates through an angle of \( \frac{\pi}{8} \) radians in one second. [2]

c. Let \( T \) seconds be the time that [AB] is watered in each revolution. Find the value of \( T \). [3]

d. Consider one clockwise revolution of the sprinkler. At \( t = 0 \), the water crosses the edge of the path at \( A \). At time \( t \) seconds, the water crosses the edge of the path at a movable point \( D \) which is a distance \( d \) meters south of point \( A \). Let \( \alpha = \angle ASD \) and \( \beta = \angle SAB \), where \( \alpha, \beta \) are measured in radians.
Sprinkler Diagram 2
Write down an expression for \( \alpha \) in terms of \( t \). [1]

e. It is known that \( \beta = 0.7754 \) radians, correct to four significant figures. By using the sine rule in \( \Delta ASD \), show that the distance, \( d \), at time \( t \), can be modeled by
\( d(t) = \frac{20 \sin\left(\frac{\pi t}{8}\right)}{\sin\left(2.37 – \frac{\pi t}{8}\right)} \). [3]

f. A turtle walks south along the edge of the path. At time \( t \) seconds, the turtle’s distance, \( g \) meters south of \( A \), can be modeled by
\( g(t) = 0.05t^2 + 1.1t + 18 \), where \( t \geq 0 \).
At \( t = 0 \), state how far south the turtle is from \( A \). [1]

g. Let \( w \) represent the distance between the turtle and point \( D \) at time \( t \) seconds.
(i) Use the expressions for \( g(t) \) and \( d(t) \) to write down an expression for \( w \) in terms of \( t \). [1]
(ii) Hence find when and where on the path the water first reaches the turtle. [3]

▶️ Answer/Explanation
Markscheme Solution

a. [3 marks]

METHOD 1:
Let \( M \) be the midpoint of [AB], so \( AB = 2AM \).
Use Pythagoras’ theorem: \( AM^2 = 20^2 – 14^2 = 400 – 196 = 204 \) (M1)
\( AM = \sqrt{204} = 2\sqrt{51} \approx 14.2828 \) (A1)
\( AB = 2 \times 14.2828 = 28.5657 \approx 28.57 \) (A1)
Answer: \( AB = 28.57 \)

METHOD 2:
Let \( \theta = \angle ASM \), \( \cos \theta = \frac{14}{20} = 0.7 \), so \( \theta = \cos^{-1}(0.7) \approx 0.795398 \) (M1)
Use trigonometry: \( AM = 20 \sin(\cos^{-1}(0.7)) = \sqrt{204} \approx 14.2828 \) (A1)
\( AB = 2 \times 14.2828 = 28.5657 \approx 28.57 \) (A1)
Answer: \( AB = 28.57 \)

b. [2 marks]

Sprinkler rotates \( 2\pi \) radians in 16 seconds (M1)
Angle per second: \( \frac{2\pi}{16} = \frac{\pi}{8} \) radians (A1)
Answer: Sprinkler rotates \( \frac{\pi}{8} \) radians/s.

c. [3 marks]

Angle \( \theta = \angle ASM \), \( \cos \theta = \frac{14}{20} = 0.7 \), so \( \theta = \cos^{-1}(0.7) \approx 0.795398 \). Total angle: \( 2 \theta = 2 \cos^{-1}(0.7) \approx 1.59079 \) (M1)
Angular speed: \( \frac{2\pi}{16} = \frac{\pi}{8} \) rad/s. Time: \( T = \frac{2 \cos^{-1}(0.7)}{\frac{\pi}{8}}} = \frac{1.59079 \cdot 8}{\pi} \approx 4.05093 \) (M1)
\( T = 4.05 \) seconds (A1)
Answer: \( \boxed{T = 4.05} \)

d. [1 mark]

Angular speed \( \frac{\pi}{8} \) rad/s, clockwise. Angle \( \alpha = \omega t = \frac{\pi t}{8} \) (A1)
Answer: \( \alpha = \frac{\pi t}{8} \)

e. [3 marks]

In \( \Delta ASD \), sine rule: \( \frac{d}{\sin \alpha} = \frac{20}{\sin \angle ADS} \) (M1)
\( \angle ADS = \pi – \beta – \alpha = \pi – 0.7754 – \alpha \approx 2.37 – \alpha \) (A1)
\( d = \frac{20 \sin \alpha}{\sin(2.37 – \alpha)} \). Substitute \( \alpha = \frac{\pi t}{8} \):
\( d(t) = \frac{20 \sin\left(\frac{\pi t}{8}\right)}{\sin\left(2.37 – \frac{\pi t}{8}\right)} \) (A1)
Answer: Shown as required.

f. [1 mark]

\( g(t) = 0.05t^2 + 1.1t + 18 \). At \( t = 0 \): \( g(0) = 18 \) (A1)
Answer: \( \boxed{18} \) meters

g. [4 marks]

(i) \( w(t) = \left| g(t) – d(t) \right| = \left| 0.05t^2 + 1.1t + 18 – \frac{20 \sin\left(\frac{\pi t}{8}\right)}{\sin\left(2.37 – \frac{\pi t}{8}\right)} \right| \) (A1)
(ii) Solve \( w(t) = 0 \): \( 0.05t^2 + 1.1t + 18 = \frac{20 \sin\left(\frac{\pi t}{8}\right)}{\sin\left(2.37 – \frac{\pi t}{8}\right)} \) (M1)
Numerical solution yields \( t = 3.34880 \approx 3.36 \) s (A1)
At \( t = 3.36 \), \( g(3.36) \approx 22.258 \approx 22.26 \) m south of \( A \) (A1)
Answer: (i) \( w(t) = \left| 0.05t^2 + 1.1t + 18 – \frac{20 \sin\left(\frac{\pi t}{8}\right)}{\sin\left(2.37 – \frac{\pi t}{8}\right)} \right| \), (ii) \( \boxed{t \approx 3.36, \, 22.26} \) meters south of \( A \).

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