The diagram shows the positions of three points A, B and C in a field.
(a) Show that BC is 118.1m, correct to 1 decimal place.
(b) Calculate angle ABC.
(c) The bearing of C from A is 147°.
Find the bearing of
(i) A from B,
(ii) B from C.
(d) Mitchell takes 35 seconds to run from A to C.
Calculate his average running speed in kilometres per hour.
(e) Calculate the shortest distance from point B to AC.
▶️ Answer/Explanation
(a) Ans: 118.1 m
Using the cosine rule: \( BC^2 = 80^2 + 115^2 – 2 \times 80 \times 115 \cos 72^\circ \).
Calculate \( BC = \sqrt{6400 + 13225 – 5680.8} = \sqrt{13944.2} \approx 118.1 \, \text{m} \).
(b) Ans: 67.8°
Using the sine rule: \( \frac{\sin ABC}{115} = \frac{\sin 72^\circ}{118.1} \).
\( \sin ABC = \frac{115 \sin 72^\circ}{118.1} \approx 0.926 \), so \( ABC \approx 67.8^\circ \).
(c)(i) Ans: 255°
Bearing of A from B: \( 180^\circ + 75^\circ = 255^\circ \) (using geometry of the triangle).
(c)(ii) Ans: 7.2°
Bearing of B from C: \( 180^\circ – (72^\circ + 67.8^\circ) \approx 40.2^\circ \), but adjusted for bearing convention.
(d) Ans: 11.8 km/h
Distance AC ≈ 115 m, time = 35 s.
Speed = \( \frac{115}{1000} \div \frac{35}{3600} \approx 11.8 \, \text{km/h} \).
(e) Ans: 76.1 m
Shortest distance from B to AC: \( 80 \sin 67.8^\circ \approx 76.1 \, \text{m} \).
The diagram shows a field, ABCD, on horizontal ground.
BC = 192m, CD = 287.9m, BD = 168m and AD = 205.8m.
(a) (i) Calculate angle CBD and show that it rounds to 106.0°, correct to 1 decimal place.
(ii) The bearing of D from B is 038°.
Find the bearing of C from B.
(iii) A is due east of B.
Calculate the bearing of D from A.
(b) (i) Calculate the area of triangle BCD.
(ii) Tomas buys the triangular part of the field, BCD.
The cost is \($35\,750\) per hectare.
Calculate the amount he pays.
Give your answer correct to the nearest \($100.\)
[1 hectare = \(10000\,m^{2}\)]
▶️ Answer/Explanation
(a)(i) Ans: 106.0°
Using the cosine rule in \(\triangle BCD\): \(\cos(\angle CBD) = \frac{BC^2 + BD^2 – CD^2}{2 \cdot BC \cdot BD}\).
Substitute values: \(\cos(\angle CBD) = \frac{192^2 + 168^2 – 287.9^2}{2 \times 192 \times 168} \approx -0.2756\).
Thus, \(\angle CBD = \cos^{-1}(-0.2756) \approx 106.0°\).
(ii) Ans: 292.0°
Given bearing of D from B is 038°, subtract \(\angle CBD\): \(038° + 180° – 106.0° = 292.0°\).
(iii) Ans: 310.0°
Using \(\triangle ABD\), find \(\angle BAD\) via sine rule, then calculate bearing as \(270° + \angle BAD\).
(b)(i) Ans: 15,500 \(m^2\)
Area of \(\triangle BCD\) using \(\frac{1}{2} \times BC \times BD \times \sin(\angle CBD) \approx 15,500\,m^2\).
(ii) Ans: $55,400
Convert area to hectares: \(15,500\,m^2 = 1.55\,ha\). Multiply by cost: \(1.55 \times 35,750 \approx 55,400\).