13.1 Trigonometry in Two and Three Dimensions

13.1 Trigonometry in Two and Three Dimensions

Quick Answer: On the calculator-free NBT MAT paper, a triangle is solved with three tools: the area formula (1/2)ab sin C, the sine rule a/sin A = b/sin B = c/sin C, and the cosine rule a² = b² + c² − 2bc cos A. Nobody labels the diagram with the name of the rule. You choose from which sides and angles are known. Two-dimensional work hides the triangle in a bearing or a pair of observation points. Three-dimensional work hides it in a pyramid or a room: drop a perpendicular to the base, then solve the right triangles that appear.

What this MAT skill is

The MAT booklet lists area, sine and cosine rules together with two-dimensional and three-dimensional applications. Independent OpenExamPrep teaching for this section uses original lengths and angles. It does not reprint confidential NBTP items, and it does not describe the resource as approved by CEA, USAf, or a university.

NSC trigonometry is often scaffolded: part (a) asks you to find an angle, part (b) says hence find a length, and a printed diagram already marks the included angle. MAT is one multiple-choice stem. The figure, if a figure is described in words, will not say included angle or opposite the 60°. You name the configuration yourself. Calculators are not allowed, so numerical work uses special angles 30°, 45°, and 60°, or exact surds.

The three tools, without a label on the sketch

Area. When two sides enclose a known angle C,

Area = (1/2) a b sin C.

C must be the included angle. If you are given two sides and a non-included angle, this formula is the wrong first move.

Sine rule.

a / sin A = b / sin B = c / sin C.

Use it when you have a matching side–opposite-angle pair and one more side or angle. Rearranged, sin A / a = sin B / b is the same statement.

Cosine rule.

a² = b² + c² − 2 b c cos A,

and the two cyclic versions for the other vertices. Use it when you have two sides and the included angle (to find the third side) or three sides (to find an angle). To find an angle:

cos A = (b² + c² − a²) / (2 b c).

If cos A is negative, A is obtuse. That sign is information, not an error.

The table is the labelling the MAT diagram will not give you.

Worked example — area, original numbers

Triangle PQR has PQ = 7, PR = 8, and included angle QPR = 60°.

Area = (1/2)(7)(8) sin 60° = 28 × (√3 / 2) = 14√3.

If a stem asked for the area and printed 28 among the options, that option used sin 60° as 1, or dropped the 1/2 and then used sin 60° = 1/2. The value 14 uses sin 60° as 1/2, which is actually sin 30°. The value 56√3 dropped the 1/2 and kept √3.

The same triangle’s third side QR comes from the cosine rule:

QR² = 7² + 8² − 2(7)(8) cos 60° = 49 + 64 − 112 × (1/2) = 113 − 56 = 57,

so QR = √57. Area and the third side are different questions; MAT will not say hence.

Worked example — sine rule

In triangle ABC, A = 30°, B = 45°, and a = 10.

First, C = 180° − 30° − 45° = 105°. That step is yours; the stem may not ask for C.

The sine-rule constant is a / sin A = 10 / sin 30° = 10 / (1/2) = 20.

Then b = 20 sin 45° = 20 × (√2 / 2) = 10√2.

And c = 20 sin 105°. Using sin 105° = sin(60° + 45°) = sin 60° cos 45° + cos 60° sin 45° = (√3/2)(√2/2) + (1/2)(√2/2) = (√6 + √2)/4, so c = 20 × (√6 + √2)/4 = 5(√6 + √2). A stem that only wanted b never mentioned 105°. You still needed to know that the sine rule uses the opposite pair, not an adjacent pair.

Worked example — cosine rule for a side and for an angle

Find a side. In triangle DEF, DE = 5, DF = 7, included angle at D is 60°. Side EF opposite D satisfies

EF² = 5² + 7² − 2(5)(7)(1/2) = 25 + 49 − 35 = 39,

so EF = √39. The trap 74 is 25 + 49 with the cosine term forgotten. The trap √74 treats the cosine term as zero. Adding the cosine term instead of subtracting produces 109, so √109. Using 2bc without the factor 1/2 subtracts 70 instead of 35 and yields EF = 2.

Find an angle. Triangle with sides 5, 7, and 8. Let E be the vertex opposite the side of length 8.

cos E = (5² + 7² − 8²) / (2 × 5 × 7) = (25 + 49 − 64) / 70 = 10 / 70 = 1/7.

The cosine is positive, so E is acute. Using 8² + 7² − 5² in the numerator would find a different angle. A negative sign on 1/7 would claim an obtuse angle that this triangle does not have at E, because 8² = 64 is less than 5² + 7² = 74 — the converse of Pythagoras says the angle opposite 8 is acute.

The ambiguous case, which a diagram will not announce

SSA: side a, angle A, and another side b. Compare a with the height h = b sin A.

• If A is acute and a < h: no triangle.
• If A is acute and a = h: one right triangle.
• If A is acute and h < a < b: two triangles (the ambiguous case).
• If A is acute and a ≥ b: one triangle.
• If A is obtuse, you need a > b for one triangle.

Worked example. a = 6, A = 30°, b = 6√2.

Height h = b sin A = 6√2 × (1/2) = 3√2. Now 3√2 < 6 < 6√2, so two triangles exist.

sin B = (b sin A) / a = (6√2 × 1/2) / 6 = √2 / 2, so B = 45° or B = 135°.

• If B = 45°, then C = 105°.
• If B = 135°, then C = 15°.

Both are consistent with A = 30°. A multiple-choice option that lists only one of the two completed triangles is incomplete. MAT will not print show that two triangles are possible.

Two-dimensional application — bearings, original numbers

A boat leaves harbour H and sails 10 km on a bearing of 030°, then 10 km on a bearing of 090°. How far is the boat from H, and on what bearing from H?

Resolve into north and east, using that a bearing is measured clockwise from north.

• First leg: north component 10 cos 30° = 10 × (√3/2) = 5√3; east component 10 sin 30° = 10 × (1/2) = 5.
• Second leg, bearing 090°: north 0, east 10.

Totals: north 5√3, east 15.

Distance from H: √( (5√3)² + 15² ) = √(75 + 225) = √300 = 10√3 km.

The angle east of north satisfies tan θ = 15 / (5√3) = 3 / √3 = √3, so θ = 60°. The bearing from H is 060°.

Nothing in the stem said use the cosine rule in triangle H–waypoint–finish. You could instead treat the two legs as sides of that triangle. The forward headings differ by 60°, but the interior angle at the waypoint is the angle between the incoming path reversed and the outgoing path, which is 180° − 60° = 120°. Then

d² = 10² + 10² − 2(10)(10) cos 120° = 200 − 200(−1/2) = 200 + 100 = 300,

d = 10√3, the same distance. Two routes, one answer. MAT will not tell you which route.

Two-dimensional application — two observers, no hence

A vertical pole TP stands on level ground. From A, the angle of elevation of T is 30° and AP = 12√3. From B, on the opposite side of P, the angle of elevation of T is 45°. Find TP and AB.

From A: tan 30° = TP / (12√3) = 1/√3, so TP = 12√3 / √3 = 12.

From B: tan 45° = 12 / BP = 1, so BP = 12.

Then AB = AP + PB = 12√3 + 12 = 12(1 + √3).

The stem can ask for AB in one sentence, with no part (a) that asked for the height. You still compute the height first. Using tan 30° as √3 would give a height 36 and an AB that is not among the designed options. Placing A and B on the same side of P would require |12√3 − 12| instead of a sum; the stem’s opposite side is the instruction that chooses the sum.

Three-dimensional application — cuboid

A room is a cuboid 3 m by 4 m by 12 m, with height 12 m. The floor is the 3-by-4 rectangle.

Floor diagonal = 5 m (the 3-4-5 triple). Space diagonal = 13 m (the 5-12-13 triple).

The angle α between the space diagonal and the floor satisfies

tan α = 12 / 5, sin α = 12 / 13, cos α = 5 / 13.

A stem that asks for tan α wants 12/5, not 5/12 (that is the tangent of the complementary angle with a vertical edge) and not 12/13 (sine, if someone treated the 13 as adjacent). No printed cube will have the floor diagonal pre-marked. You draw the floor diagonal yourself, then the right triangle whose legs are that diagonal and the height.

Three-dimensional application — square pyramid

Square base ABCD with AB = 6. Apex V, with the foot of the perpendicular from V to the base at the centre O, and VO = 4.

OA is half of the base diagonal: base diagonal = 6√2, so OA = 3√2.

Slant edge VA = √(VO² + OA²) = √(16 + 18) = √34.

The angle of inclination of VA to the base is the angle between VA and OA, and

tan θ = 4 / (3√2) = (4√2) / 6 = (2√2) / 3.

The angle between a triangular face, say VAB, and the base is different. Let M be the midpoint of AB. Then OM = 3 (the distance from the centre of a square of side 6 to a side), and

tan φ = VO / OM = 4/3.

Confusing θ with φ is the 3D trap: an edge inclination is not a face inclination. MAT will not caption the diagram angle between the face and the base. Using √34 as a perpendicular height in a later volume question would also be wrong; that length is a slant edge.

Method on the afternoon paper

1. Write every given length and angle. Mark included versus opposite yourself.
2. If two sides enclose a known angle, area or cosine rule. If a side is paired with its opposite angle, sine rule. If three sides, cosine rule.
3. For SSA, compute the height before you commit to one triangle.
4. In 2D stories, draw north–east components or two right triangles that share a height.
5. In 3D stories, drop a perpendicular to the base, name the base point, and only then apply 2D trigonometry.
6. Use exact special-angle values. There is no calculator to hide a wrong ratio.

Traps

• Using (1/2)ab sin C when C is not the included angle
• Pairing a side with an adjacent angle in the sine rule
• Forgetting the cosine-rule minus sign, or using cos 60° as √3/2
• Listing only one SSA triangle when h < a < b
• Taking the triangle angle at a waypoint as the heading change instead of its supplement
• Mixing edge inclination with face inclination in a pyramid
• Reaching for a calculator; MAT is calculator-free

Independent OpenExamPrep practice in this section is that unlabelled choice of rule, with original 7-8-60 triangles, 10 km bearings, a 3-4-12 room, and a square pyramid of base 6 and height 4.