Year 12 · Measurement
Quick tips — memory joggers
Sine rule
Cosine rule
Area of a triangle
Bearings & applications
Level 1 · Fluency
In a right-angled triangle, which ratio equals \(\dfrac{\text{opposite}}{\text{adjacent}}\)?
The tangent ratio (\(\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}}\)).
In a right-angled triangle, which ratio equals \(\dfrac{\text{opposite}}{\text{hypotenuse}}\)?
The sine ratio (\(\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}\)).
In a right-angled triangle, which ratio equals \(\dfrac{\text{adjacent}}{\text{hypotenuse}}\)?
The cosine ratio (\(\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}\)).
Level 2 · Application
For the right-angled triangle shown, find the size of the angle at B, correct to the nearest degree.
\(37^{\circ}\)
\(\tan B = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{9}{12}\), so \(B = \tan^{-1}(0.75) =\) \(37^{\circ}\).
A right-angled triangle has its right angle at \(C\). The side opposite angle \(B\) is 5 cm and the side adjacent to \(B\) is 12 cm. Find angle \(B\), to the nearest degree.
\(23^{\circ}\)
\(\tan B = \dfrac{5}{12}\), so \(B = \tan^{-1}(0.4167) =\) \(23^{\circ}\).
A right-angled triangle has its right angle at \(C\). The side opposite angle \(A\) is 8 cm and the hypotenuse is 17 cm. Find angle \(A\), to the nearest degree.
\(28^{\circ}\)
\(\sin A = \dfrac{8}{17} = 0.4706\), so \(A = \sin^{-1}(0.4706) =\) \(28^{\circ}\).
Level 3 · Further Application
In the right-angled triangle shown (right angle at C, with the two shorter sides 9 and 12):
15
(a) \(\tan B = \dfrac{9}{12} \Rightarrow B = 37^{\circ}\).
(b) \(AB = \sqrt{9^{2} + 12^{2}} = \sqrt{225} =\) 15.
A right-angled triangle has its right angle at \(C\), with the two shorter sides 5 cm (opposite \(B\)) and 12 cm (opposite \(A\)).
\(23^{\circ}\); 13 cm
(a) \(\tan B = \dfrac{5}{12} \Rightarrow B =\) \(23^{\circ}\).
(b) \(AB = \sqrt{5^{2} + 12^{2}} = \sqrt{169} =\) 13 cm.
A ladder leans against a wall. Its foot is 2.5 m from the wall and it reaches 6 m up the wall.
\(67^{\circ}\); 6.5 m
(a) \(\tan\theta = \dfrac{6}{2.5} = 2.4 \Rightarrow \theta = \tan^{-1}(2.4) =\) \(67^{\circ}\).
(b) length \(= \sqrt{2.5^{2} + 6^{2}} = \sqrt{42.25} =\) 6.5 m.
Level 1 · Fluency
Write the formula for the area of a triangle given two sides \(a\), \(b\) and their included angle \(C\).
\(A = \tfrac{1}{2}ab\sin C\).
In the area formula \(A = \tfrac{1}{2}ab\sin C\), what does the angle \(C\) represent?
The angle included (enclosed) between the two sides \(a\) and \(b\).
State the formula for the area of a triangle with sides \(p\) and \(q\) and included angle \(R\).
\(A = \tfrac{1}{2}pq\sin R\).
Level 2 · Application
A triangle has sides 7 cm and 9 cm with an included angle of \(65^{\circ}\). Calculate its area, to two decimal places.
28.55 cm²
\(A = \tfrac{1}{2} \times 7 \times 9 \times \sin 65^{\circ} =\) 28.55 cm².
A triangle has sides 12 cm and 8 cm with an included angle of \(40^{\circ}\). Calculate its area, to two decimal places.
30.85 cm²
\(A = \tfrac{1}{2} \times 12 \times 8 \times \sin 40^{\circ} = 48\sin 40^{\circ} =\) 30.85 cm².
A triangle has sides 15 cm and 6 cm with an included angle of \(108^{\circ}\). Calculate its area, to two decimal places.
42.80 cm²
\(A = \tfrac{1}{2} \times 15 \times 6 \times \sin 108^{\circ} = 45\sin 108^{\circ} =\) 42.80 cm².
Level 3 · Further Application
A triangle \(OAB\) has \(OA = OB = 10\) cm and the angle \(AOB = 30^{\circ}\).
25 cm²; 300 cm²
(a) \(A = \tfrac{1}{2} \times 10 \times 10 \times \sin 30^{\circ} =\) 25 cm².
(b) \(12 \times 25 =\) 300 cm².
A triangle \(OAB\) has \(OA = OB = 8\) cm and the angle \(AOB = 45^{\circ}\).
22.63 cm²; 181.02 cm²
(a) \(A = \tfrac{1}{2} \times 8 \times 8 \times \sin 45^{\circ} = 32\sin 45^{\circ} =\) 22.63 cm².
(b) \(8 \times 22.627 =\) 181.02 cm².
A triangle has an area of 20 cm², with two sides 8 cm and 10 cm enclosing an angle \(C\).
\(\sin C = 0.5\); \(C = 30^{\circ}\)
(a) \(20 = \tfrac{1}{2} \times 8 \times 10 \times \sin C = 40\sin C \Rightarrow \sin C =\) \(0.5\).
(b) \(C = \sin^{-1}(0.5) =\) \(30^{\circ}\) (the acute value).
Level 1 · Fluency
Write the sine rule relating sides \(a\), \(b\) and angles \(A\), \(B\).
\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B}\).
Write the sine rule in the form used to find an unknown angle (sines over sides).
\(\dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}\).
In triangle \(ABC\), which side is opposite angle \(A\)?
Side \(a\) (the side opposite a vertex is named with the matching lower-case letter).
Level 2 · Application
In triangle \(ABC\), angle \(A = 40^{\circ}\), angle \(B = 75^{\circ}\) and side \(a = 10\) cm. Find side \(b\), to one decimal place.
15.0 cm
\(b = \dfrac{a\sin B}{\sin A} = \dfrac{10 \times \sin 75^{\circ}}{\sin 40^{\circ}} =\) 15.0 cm.
In triangle \(ABC\), angle \(A = 35^{\circ}\), angle \(C = 80^{\circ}\) and side \(a = 12\) cm. Find side \(c\), to one decimal place.
20.6 cm
\(c = \dfrac{a\sin C}{\sin A} = \dfrac{12 \times \sin 80^{\circ}}{\sin 35^{\circ}} =\) 20.6 cm.
In triangle \(ABC\), angle \(B = 50^{\circ}\), angle \(C = 60^{\circ}\) and side \(b = 9\) cm. Find side \(c\), to one decimal place.
10.2 cm
\(c = \dfrac{b\sin C}{\sin B} = \dfrac{9 \times \sin 60^{\circ}}{\sin 50^{\circ}} =\) 10.2 cm.
Level 3 · Further Application
In triangle \(PQR\), angle \(P = 52^{\circ}\), angle \(Q = 61^{\circ}\) and side \(p\) (opposite \(P\)) \(= 14\) cm.
16.4 cm
(a) \(R = 180 - 52 - 61 = 67^{\circ}\).
(b) \(r = \dfrac{14 \times \sin 67^{\circ}}{\sin 52^{\circ}} =\) 16.4 cm.
In triangle \(XYZ\), angle \(X = 48^{\circ}\), angle \(Y = 57^{\circ}\) and side \(x\) (opposite \(X\)) \(= 20\) cm.
\(Z = 75^{\circ}\); \(z = 26.0\) cm
(a) \(Z = 180^{\circ} - 48^{\circ} - 57^{\circ} = 75^{\circ}\).
(b) \(z = \dfrac{20 \times \sin 75^{\circ}}{\sin 48^{\circ}} =\) 26.0 cm.
In triangle \(ABC\), angle \(A = 43^{\circ}\), angle \(B = 68^{\circ}\) and side \(a\) (opposite \(A\)) \(= 15\) cm.
\(C = 69^{\circ}\); \(b = 20.4\) cm
(a) \(C = 180^{\circ} - 43^{\circ} - 68^{\circ} = 69^{\circ}\).
(b) \(b = \dfrac{15 \times \sin 68^{\circ}}{\sin 43^{\circ}} =\) 20.4 cm.
Level 1 · Fluency
The sine rule gives \(\sin A = 0.6\) and the triangle’s angle \(A\) is known to be obtuse. Find \(A\), to the nearest degree.
\(143^{\circ}\)
\(A = 180^{\circ} - \sin^{-1}(0.6) =\) \(143^{\circ}\).
The sine rule gives \(\sin\theta = 0.5\) and \(\theta\) is obtuse. Find \(\theta\).
\(150^{\circ}\)
\(\theta = 180^{\circ} - \sin^{-1}(0.5) = 180^{\circ} - 30^{\circ} =\) \(150^{\circ}\).
The sine rule gives \(\sin A = 0.8\) and \(A\) is obtuse. Find \(A\), to the nearest degree.
\(127^{\circ}\)
\(A = 180^{\circ} - \sin^{-1}(0.8) = 180^{\circ} - 53^{\circ} =\) \(127^{\circ}\).
Level 2 · Application
In a triangle, \(\sin\theta = 0.68\) and \(\theta\) is obtuse. Find \(\theta\), to the nearest degree.
\(137^{\circ}\)
\(\theta = 180^{\circ} - \sin^{-1}(0.68) =\) \(137^{\circ}\).
In a triangle, \(\sin\theta = 0.42\) and \(\theta\) is obtuse. Find \(\theta\), to the nearest degree.
\(155^{\circ}\)
\(\theta = 180^{\circ} - \sin^{-1}(0.42) = 180^{\circ} - 25^{\circ} =\) \(155^{\circ}\).
In a triangle, \(\sin\theta = 0.91\) and \(\theta\) is obtuse. Find \(\theta\), to the nearest degree.
\(114^{\circ}\)
\(\theta = 180^{\circ} - \sin^{-1}(0.91) = 180^{\circ} - 66^{\circ} =\) \(114^{\circ}\).
Level 3 · Further Application
In triangle \(ABC\), \(AC = 22\) cm, \(BC = 14\) cm, angle \(BAC = 32^{\circ}\), and angle \(ABC\) is obtuse.
\(124^{\circ}\)
(a) \(\dfrac{\sin(ABC)}{22} = \dfrac{\sin 32^{\circ}}{14} \Rightarrow \sin(ABC) = \dfrac{22 \times \sin 32^{\circ}}{14} = 0.8327\).
(b) \(ABC\) is obtuse: \(ABC = 180^{\circ} - \sin^{-1}(0.8327) =\) \(124^{\circ}\).
In triangle \(ABC\), \(AB = 18\) cm, \(BC = 11\) cm, angle \(BAC = 35^{\circ}\), and angle \(ACB\) is obtuse.
\(110^{\circ}\)
(a) \(AB\) is opposite \(ACB\) and \(BC\) is opposite \(BAC\), so \(\dfrac{\sin(ACB)}{18} = \dfrac{\sin 35^{\circ}}{11} \Rightarrow \sin(ACB) = \dfrac{18 \times \sin 35^{\circ}}{11} = 0.9386\).
(b) \(ACB\) is obtuse: \(ACB = 180^{\circ} - \sin^{-1}(0.9386) =\) \(110^{\circ}\).
In triangle \(PQR\), \(PR = 25\) cm, \(QR = 16\) cm, angle \(QPR = 30^{\circ}\), and angle \(PQR\) is obtuse.
\(129^{\circ}\)
(a) \(PR\) is opposite \(PQR\) and \(QR\) is opposite \(QPR\), so \(\dfrac{\sin(PQR)}{25} = \dfrac{\sin 30^{\circ}}{16} \Rightarrow \sin(PQR) = \dfrac{25 \times \sin 30^{\circ}}{16} = 0.7813\).
(b) \(PQR\) is obtuse: \(PQR = 180^{\circ} - \sin^{-1}(0.7813) =\) \(129^{\circ}\).
Level 1 · Fluency
Write the cosine rule for side \(c\) in terms of \(a\), \(b\) and angle \(C\).
\(c^{2} = a^{2} + b^{2} - 2ab\cos C\).
Rearrange the cosine rule to make \(\cos C\) the subject.
\(\cos C = \dfrac{a^{2}+b^{2}-c^{2}}{2ab}\).
Which rule should you use to find the third side of a triangle given two sides and the angle between them?
The cosine rule (this is the SAS case — two sides and the included angle).
Level 2 · Application
A triangle has sides \(a = 7\) cm and \(b = 9\) cm with an included angle \(C = 65^{\circ}\). Find side \(c\), to one decimal place.
8.8 cm
\(c^{2} = 7^{2} + 9^{2} - 2 \times 7 \times 9 \times \cos 65^{\circ} = 76.75\); \(c =\) 8.8 cm.
A triangle has sides \(a = 10\) cm and \(b = 6\) cm with an included angle \(C = 50^{\circ}\). Find side \(c\), to one decimal place.
7.7 cm
\(c^{2} = 10^{2} + 6^{2} - 2 \times 10 \times 6 \times \cos 50^{\circ} = 58.87\); \(c =\) 7.7 cm.
A triangle has sides \(a = 13\) cm and \(b = 8\) cm with an included angle \(C = 110^{\circ}\). Find side \(c\), to one decimal place.
17.4 cm
\(c^{2} = 13^{2} + 8^{2} - 2 \times 13 \times 8 \times \cos 110^{\circ} = 304.14\); \(c =\) 17.4 cm.
Level 3 · Further Application
Two hikers leave a point \(P\): one walks 8 km on a bearing of \(040^{\circ}\), the other 11 km on a bearing of \(115^{\circ}\). The angle between their paths at \(P\) is \(75^{\circ}\).
11.8 km
(a) \(d^{2} = 8^{2} + 11^{2} - 2 \times 8 \times 11 \times \cos 75^{\circ} \Rightarrow d =\) 11.8 km.
(b) We know two sides and the included angle, which is exactly the case the cosine rule handles (the sine rule needs an angle opposite a known side).
Two roads leave a town \(T\): one runs 15 km to a farm \(F\), the other 9 km to a dam \(D\), and the angle \(FTD\) between the roads is \(50^{\circ}\).
11.5 km
(a) \(FD^{2} = 15^{2} + 9^{2} - 2 \times 15 \times 9 \times \cos 50^{\circ} = 132.45 \Rightarrow FD =\) 11.5 km.
(b) Two sides and the included angle are known (SAS), so the cosine rule applies directly.
A triangular field \(ABC\) has \(AB = 40\) m, \(AC = 55\) m and the included angle \(BAC = 60^{\circ}\).
49 m
(a) \(BC^{2} = 40^{2} + 55^{2} - 2 \times 40 \times 55 \times \cos 60^{\circ} = 2425 \Rightarrow BC =\) 49 m.
(b) We have two sides and the included angle (SAS); the sine rule would need an angle opposite a known side, which we do not have.
Level 1 · Fluency
A true bearing is measured clockwise from which direction?
From north (\(000^{\circ}\)), measured clockwise, using three digits.
How many digits are used to write a true bearing?
Three digits, e.g. \(072^{\circ}\) or \(200^{\circ}\).
A compass bearing such as N30°E begins from which direction?
From the nearer of north or south (here north), then turning the stated number of degrees towards east or west.
Level 2 · Application
Convert the compass bearing S40°E to a true bearing.
\(140^{\circ}\)
Start at south (\(180^{\circ}\)) and rotate \(40^{\circ}\) toward east (anticlockwise from south is... measure): S40°E \(= 180^{\circ} - 40^{\circ} =\) \(140^{\circ}\).
Convert the compass bearing N70°E to a true bearing.
\(070^{\circ}\)
Start at north (\(000^{\circ}\)) and rotate \(70^{\circ}\) clockwise toward east: N70°E \(=\) \(070^{\circ}\).
Convert the compass bearing S25°W to a true bearing.
\(205^{\circ}\)
Start at south (\(180^{\circ}\)) and rotate \(25^{\circ}\) clockwise toward west: \(180^{\circ} + 25^{\circ} =\) \(205^{\circ}\).
Level 3 · Further Application
A hiker walks on a true bearing of \(295^{\circ}\).
N65°W
(a) \(295^{\circ}\) is between \(270^{\circ}\) (W) and \(360^{\circ}\) (N), so the north-west quadrant.
(b) It is \(360^{\circ} - 295^{\circ} = 65^{\circ}\) west of north, i.e. N65°W.
A ship travels on a true bearing of \(225^{\circ}\).
S45°W
(a) \(225^{\circ}\) is between \(180^{\circ}\) (S) and \(270^{\circ}\) (W), so the south-west quadrant.
(b) It is \(225^{\circ} - 180^{\circ} = 45^{\circ}\) west of south, i.e. S45°W.
A plane flies on a true bearing of \(160^{\circ}\).
S20°E
(a) \(160^{\circ}\) is between \(90^{\circ}\) (E) and \(180^{\circ}\) (S), so the south-east quadrant.
(b) It is \(180^{\circ} - 160^{\circ} = 20^{\circ}\) east of south, i.e. S20°E.
Level 1 · Fluency
If the bearing of B from A is \(070^{\circ}\), what is the bearing of A from B?
\(250^{\circ}\)
Add \(180^{\circ}\): \(070^{\circ} + 180^{\circ} =\) \(250^{\circ}\).
The bearing of Q from P is \(130^{\circ}\). What is the bearing of P from Q?
\(310^{\circ}\)
Add \(180^{\circ}\): \(130^{\circ} + 180^{\circ} =\) \(310^{\circ}\).
The bearing of Y from X is \(200^{\circ}\). What is the bearing of X from Y?
\(020^{\circ}\)
The bearing exceeds \(180^{\circ}\), so subtract \(180^{\circ}\): \(200^{\circ} - 180^{\circ} =\) \(020^{\circ}\).
Level 2 · Application
A ship sails 12 km due east then 5 km due north. Find its bearing from the start, to the nearest degree.
\(067^{\circ}\)
Angle east-of-north \(= \tan^{-1}\left(\dfrac{12}{5}\right) = 67^{\circ}\), so the bearing is about \(067^{\circ}\) (N67°E).
A ship sails 6 km due north then 8 km due east. Find its bearing from the start, to the nearest degree.
\(053^{\circ}\)
Angle east-of-north \(= \tan^{-1}\left(\dfrac{8}{6}\right) = 53^{\circ}\), so the bearing is about \(053^{\circ}\) (N53°E).
A hiker walks 10 km due south then 4 km due west. Find the bearing of the finish from the start, to the nearest degree.
\(202^{\circ}\)
The finish is in the SW quadrant. Angle west-of-south \(= \tan^{-1}\left(\dfrac{4}{10}\right) = 22^{\circ}\), so the bearing is \(180^{\circ} + 22^{\circ} =\) \(202^{\circ}\) (S22°W).
Level 3 · Further Application
From a camp \(P\), group A walks on a bearing of \(040^{\circ}\) and group B on a bearing of \(115^{\circ}\). After finding \(AB\) with the cosine rule, the bearing of B from A is required.
(a) With \(AB\) known (from the cosine rule) and \(PB\) and the angle at \(P\) known, the sine rule gives angle \(PAB\): \(\dfrac{\sin(PAB)}{PB} = \dfrac{\sin P}{AB}\).
(b) The bearing of B from A is found by adding angle \(PAB\) to the reverse bearing of P from A (\(220^{\circ}\)), giving the required true bearing.
A yacht sails 20 km on a bearing of \(060^{\circ}\) from harbour \(H\) to buoy \(A\), then 15 km on a bearing of \(150^{\circ}\) to buoy \(B\).
25.0 km
(a) At \(A\), the reverse of the incoming leg is \(060^{\circ} + 180^{\circ} = 240^{\circ}\); the outgoing leg is \(150^{\circ}\). The angle between them is \(240^{\circ} - 150^{\circ} = 90^{\circ}\), so \(HAB = 90^{\circ}\).
(b) With a right angle at \(A\), \(HB = \sqrt{20^{2} + 15^{2}} = \sqrt{625} =\) 25.0 km.
From port \(P\), a boat sails 30 km on a bearing of \(070^{\circ}\) to point \(Q\), then turns and sails 40 km on a bearing of \(160^{\circ}\) to point \(R\).
50.0 km
(a) At \(Q\), the reverse of the incoming leg is \(070^{\circ} + 180^{\circ} = 250^{\circ}\); the outgoing leg is \(160^{\circ}\). The angle between them is \(250^{\circ} - 160^{\circ} = 90^{\circ}\), so \(PQR = 90^{\circ}\).
(b) With a right angle at \(Q\), \(PR = \sqrt{30^{2} + 40^{2}} = \sqrt{2500} =\) 50.0 km.
Level 1 · Fluency
Name one natural feature traditionally used for navigation without instruments.
Any of: the stars/constellations, the sun, ocean swells, winds, or landmarks.
Name one celestial body traditionally used to find direction during the day.
The sun — its rising point (east) and setting point (west) give direction.
Name one environmental cue, other than the sky, that sailors have used to hold a course.
Ocean swells (or waves), or the direction of a steady prevailing wind.
Level 2 · Application
Describe one way Aboriginal and Torres Strait Islander peoples have traditionally used the night sky to navigate or mark direction.
Star patterns (for example, using particular stars or the position of constellations as they rise and set) were used to hold a direction and to mark seasons and routes across Country.
Describe how the rising and setting points of the sun can help a traveller keep a constant direction.
The sun rises roughly in the east and sets roughly in the west. By keeping the rising (or setting) sun at a fixed position relative to the direction of travel — for example, over the right shoulder — a traveller can hold a roughly constant heading through the day.
Explain how knowledge of the seasons and star positions could help both time and direct a journey across Country.
Particular stars and constellations appear at set times of year and rise and set at known points on the horizon. Recognising them lets people choose the right season to travel (when food and water are available) and follow a consistent direction by aligning their path with those stars.
Level 3 · Further Application
Traditional navigation often used the environment rather than a compass.
(a) e.g. the rising/setting position of the sun or particular stars, and the direction of prevailing winds or ocean swells.
(b) A true bearing gives a precise, repeatable number (to the degree) that can be recorded and communicated exactly, independent of weather or time of day.
Star-based navigation depends on being able to see the sky.
(a) Cloud cover can hide the stars, and stars can only be used at night (not during daylight). (b) A compass works day or night and in most weather, giving a usable direction at any time.
Ocean swells were used by island navigators to hold a heading.
(a) Long ocean swells travel in consistent directions set by distant weather systems; by feeling the regular rise and fall of the boat and keeping it at a fixed angle to the swell, a navigator can hold a steady heading. (b) Swell directions can shift with the weather and are hard to measure exactly, whereas a true bearing is a precise, repeatable number.
Level 1 · Fluency
From a point, the angle of elevation to the top of a tower is measured. Is this angle measured above or below the horizontal?
Above the horizontal (an angle of elevation looks upward).
An angle of depression is measured from the horizontal in which direction?
Below the horizontal (looking downward).
In a right-angled triangle, which theorem finds the third side when the two other sides are known and no angle is given?
Pythagoras’ theorem (\(c^{2} = a^{2} + b^{2}\)).
Level 2 · Application
From 40 m away, the angle of elevation to the top of a building is \(32^{\circ}\). Find the building’s height, to one decimal place.
25.0 m
height \(= 40 \times \tan 32^{\circ} =\) 25.0 m.
From 60 m away, the angle of elevation to the top of a tower is \(28^{\circ}\). Find the tower’s height, to one decimal place.
31.9 m
height \(= 60 \times \tan 28^{\circ} =\) 31.9 m.
A 50 m tall lighthouse is viewed from a boat. The angle of elevation to its top is \(18^{\circ}\). Find the horizontal distance from the boat to the lighthouse, to the nearest metre.
154 m
\(\tan 18^{\circ} = \dfrac{50}{d} \Rightarrow d = \dfrac{50}{\tan 18^{\circ}} =\) 154 m.
Level 3 · Further Application
A boat is 300 m from the base of a cliff. From the boat, the angle of elevation to the cliff top is \(22^{\circ}\).
121 m
(a) height \(= 300 \times \tan 22^{\circ} =\) 121 m.
(b) The angle of depression equals the angle of elevation (\(22^{\circ}\)) — they are alternate angles between the horizontal lines at the boat and the cliff top.
A drone hovers 80 m above level ground. From the drone, the angle of depression to a marker on the ground is \(35^{\circ}\).
114 m; 139 m
The angle of depression (\(35^{\circ}\)) equals the angle of elevation at the marker.
(a) \(\tan 35^{\circ} = \dfrac{80}{d} \Rightarrow d = \dfrac{80}{\tan 35^{\circ}} =\) 114 m.
(b) straight-line \(= \dfrac{80}{\sin 35^{\circ}} =\) 139 m.
From the top of a 45 m cliff, the angle of depression to a boat is \(27^{\circ}\).
88 m; 80 m
The angle of depression equals the angle of elevation at the boat.
(a) \(d_{1} = \dfrac{45}{\tan 27^{\circ}} =\) 88 m.
(b) \(d_{2} = \dfrac{45}{\tan 15^{\circ}} = 168\) m, so the boat moved \(168 - 88 =\) 80 m.
Level 1 · Fluency
How many minutes are in one degree?
60 minutes (\(1^{\circ} = 60'\)).
Write \(0.25^{\circ}\) in minutes.
\(15'\)
\(0.25 \times 60 =\) \(15'\).
Write \(45'\) as a decimal number of degrees.
\(0.75^{\circ}\)
\(45 \div 60 =\) \(0.75^{\circ}\).
Level 2 · Application
An angle is calculated as \(37.6^{\circ}\). Express it in degrees and minutes.
\(37^{\circ}36'\)
\(0.6^{\circ} = 0.6 \times 60 = 36'\), so \(37.6^{\circ} =\) \(37^{\circ}36'\).
An angle is calculated as \(52.4^{\circ}\). Express it in degrees and minutes.
\(52^{\circ}24'\)
\(0.4^{\circ} = 0.4 \times 60 = 24'\), so \(52.4^{\circ} =\) \(52^{\circ}24'\).
An angle is calculated as \(18.85^{\circ}\). Express it in degrees and minutes, to the nearest minute.
\(18^{\circ}51'\)
\(0.85^{\circ} = 0.85 \times 60 = 51'\), so \(18.85^{\circ} =\) \(18^{\circ}51'\).
Level 3 · Further Application
An angle is found from \(\tan\theta = 0.75\).
\(36^{\circ}52'\)
(a) \(\theta = \tan^{-1}(0.75) = 36.87^{\circ}\).
(b) The \(0.87^{\circ}\) part is \(52'\), so \(\theta \approx\) \(36^{\circ}52'\).
An angle is found from \(\cos\theta = 0.62\).
\(51^{\circ}41'\)
(a) \(\theta = \cos^{-1}(0.62) = 51.68^{\circ}\).
(b) \(0.68^{\circ} = 0.68 \times 60 \approx 41'\), so \(\theta \approx\) \(51^{\circ}41'\).
An angle is found from \(\sin\theta = 0.34\).
\(19^{\circ}53'\)
(a) \(\theta = \sin^{-1}(0.34) = 19.88^{\circ}\).
(b) \(0.88^{\circ} = 0.88 \times 60 \approx 53'\), so \(\theta \approx\) \(19^{\circ}53'\).
Level 1 · Fluency
In a compass radial survey, all measurements are taken from which point?
A single central point (the station), from which bearings and distances to each corner are measured.
In a radial survey, what two quantities are recorded from the central station to each corner?
The bearing (direction) and the distance to each corner.
Into what shape are the sectors of a radial survey divided so that their areas can be calculated?
Triangles — each having the central station as one vertex.
Level 2 · Application
In a radial survey from O, \(OB = 25\) m and \(OC = 30\) m, with the angle \(BOC = 70^{\circ}\). Find the area of triangle \(BOC\), to the nearest square metre.
352 m²
\(A = \tfrac{1}{2} \times 25 \times 30 \times \sin 70^{\circ} =\) 352 m².
In a radial survey from O, \(OA = 20\) m and \(OB = 28\) m, with the angle \(AOB = 55^{\circ}\). Find the area of triangle \(AOB\), to the nearest square metre.
229 m²
\(A = \tfrac{1}{2} \times 20 \times 28 \times \sin 55^{\circ} = 280\sin 55^{\circ} =\) 229 m².
In a radial survey from O, \(OP = 32\) m and \(OQ = 18\) m, with the angle \(POQ = 84^{\circ}\). Find the area of triangle \(POQ\), to the nearest square metre.
286 m²
\(A = \tfrac{1}{2} \times 32 \times 18 \times \sin 84^{\circ} = 288\sin 84^{\circ} =\) 286 m².
Level 3 · Further Application
In a radial survey from O, triangle \(COB\) has \(OC = 30\) m, \(OB = 24\) m and area 324 m².
\(64^{\circ}\); 29.1 m
(a) \(324 = \tfrac{1}{2} \times 30 \times 24 \times \sin(COB) \Rightarrow \sin(COB) = \dfrac{648}{720} = 0.9000 \Rightarrow COB =\) \(64^{\circ}\).
(b) \(CB^{2} = 30^{2} + 24^{2} - 2 \times 30 \times 24 \times \cos 64^{\circ} \Rightarrow CB =\) 29.1 m.
In a radial survey from O, triangle \(AOB\) has \(OA = 26\) m, \(OB = 20\) m and area 180 m².
\(44^{\circ}\); 18.1 m
(a) \(180 = \tfrac{1}{2} \times 26 \times 20 \times \sin(AOB) \Rightarrow \sin(AOB) = \dfrac{180}{260} = 0.6923 \Rightarrow AOB =\) \(44^{\circ}\).
(b) \(AB^{2} = 26^{2} + 20^{2} - 2 \times 26 \times 20 \times \cos 44^{\circ} = 327.9 \Rightarrow AB =\) 18.1 m.
In a radial survey from O, triangle \(POQ\) has \(OP = 40\) m, \(OQ = 25\) m and area 400 m².
\(53^{\circ}\); 32.0 m
(a) \(400 = \tfrac{1}{2} \times 40 \times 25 \times \sin(POQ) \Rightarrow \sin(POQ) = \dfrac{400}{500} = 0.8000 \Rightarrow POQ =\) \(53^{\circ}\).
(b) \(PQ^{2} = 40^{2} + 25^{2} - 2 \times 40 \times 25 \times \cos 53^{\circ} = 1021.4 \Rightarrow PQ =\) 32.0 m.