IB Mathematics SL 3.3 Applications of trigonometry AA SL Paper 2- Exam Style Questions- New Syllabus

(a) Distance \( JL \)

From the diagram, \( \angle JCL = 28^\circ \) (since bearing 062° gives 62° from north, and \( CL \) is east-west).

Using cosine rule in triangle \( JCL \):

\( JL^2 = JC^2 + CL^2 – 2(JC)(CL)\cos(\angle JCL) \)

\( JL^2 = (5.5)^2 + 8^2 – 2(5.5)(8)\cos 28^\circ \)

\( JL^2 = 30.25 + 64 – 88\cos 28^\circ \)

\( JL^2 = 94.25 – 77.685… = 16.564… \)

\( JL = \sqrt{16.564…} = 4.068… \)

\( \boxed{4.07 \text{ km}} \) (to 3 significant figures)

(b) Distance \( BL \)

Using the angle of depression of \( 0.94^\circ \), we consider a right triangle with:

• Height of lighthouse = 60 m
• Angle at lighthouse top = \( 0.94^\circ \) (angle between horizontal and line to \( B \))

\( \tan(0.94^\circ) = \frac{60}{BL} \)

\( BL = \frac{60}{\tan(0.94^\circ)} \)

\( BL = \frac{60}{0.01641…} = 3656.85… \)

\( \boxed{3660 \text{ m}} \) (to 3 significant figures) or \( 3.66 \text{ km} \)

(c) Bearing from \( L \) to \( B \)

If bearing from \( B \) to \( L \) is \( 121^\circ \), then bearing from \( L \) to \( B \) is the back-bearing:

Back-bearing = \( 121^\circ + 180^\circ = 301^\circ \) (or equivalently N59°W)

\( \boxed{301^\circ} \)

(d) Time for lighthouse rescue boat

Distance \( BL = 3.6568… \text{ km} \)
Speed = \( 48 \text{ km h}^{-1} \)

\( \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{3.6568…}{48} = 0.076184… \text{ hours} \)

Convert to minutes: \( 0.076184… \times 60 = 4.571… \text{ minutes} \)

\( \boxed{4.57 \text{ minutes}} \) (to 3 significant figures)

(e) Comparing rescue boats

Coastguard boat:
Need distance \( BC \). In triangle \( BCL \), we have:
• \( BL = 3.6568… \text{ km} \)
• \( CL = 8 \text{ km} \)
• \( \angle BLC = 31^\circ \) (from geometry: \( 90^\circ – 59^\circ \))

Using cosine rule:

\( BC^2 = BL^2 + CL^2 – 2(BL)(CL)\cos(\angle BLC) \)

\( BC^2 = (3.6568…)^2 + 8^2 – 2(3.6568…)(8)\cos 31^\circ \)

\( BC^2 = 13.372… + 64 – 58.508…\cos 31^\circ \)

\( BC^2 = 77.372… – 50.152… = 27.220… \)

\( BC = \sqrt{27.220…} = 5.217… \text{ km} \)

Time for coastguard boat: \( \frac{5.217…}{55} = 0.094859… \text{ hours} = 5.691… \text{ minutes} \)

Comparing times:
• Lighthouse boat: 4.571 minutes
• Coastguard boat: 5.691 minutes

The lighthouse rescue boat arrives first since \( 4.571 < 5.691 \).