GELE Mathematics — Plane and Spherical TrigonometryExam Answer Templates

Marks

1

Topic

Trigonometric Identities

Difficulty

easy

Template Id

T1

Examiner Tip

This is a pure recall item. Write the equation cleanly. Do not waste time explaining it unless the question asks for a derivation.

Model Answer

The Pythagorean identity states: sin²θ + cos²θ = 1 where θ is any angle.

Question Type

very_short_answer

Answer Structure

• Line 1: Write the identity equation correctly — sin²θ + cos²θ = 1 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing sinθ + cosθ = 1 (omitting the squares) — zero marks.
• Writing sin²θ − cos²θ = 1 (wrong sign) — zero marks.
• Confusing with the cofunction identity — zero marks.

Key Phrases To Include

• sin²θ + cos²θ = 1
• Pythagorean identity

Marks

1

Topic

Double-Angle Identities

Difficulty

easy

Template Id

T2

Examiner Tip

Even for a 1-mark question, write the formula line. It costs 5 seconds and protects partial credit if there is an arithmetic slip.

Model Answer

Using the double-angle formula: sin 2θ = 2 sin θ cos θ sin 2(30°) = 2 sin 30° cos 30° = 2 (0.5)(0.8660) = 0.866 ∴ sin 60° = 0.866

Question Type

numerical

Answer Structure

• Line 1: State the formula sin 2θ = 2 sin θ cos θ [implied method mark]
• Line 2: Substitute θ = 30° with known values [1 mark for correct final value]

Scoring Breakdown

Common Mark Deductions

• Giving sin 30° = 0.866 instead of sin 60° — wrong answer, zero marks.
• Writing sin 2(30°) = 2(30°) sin without the cosine factor — wrong formula, zero marks.

Key Phrases To Include

• sin 2θ = 2 sin θ cos θ
• 0.866
• √3/2

Marks

2

Topic

Law of Sines — Oblique Triangles

Difficulty

easy

Template Id

T3

Examiner Tip

Examiners are specifically testing whether you know the SSA ambiguous case — always mention it when listing Law of Sines conditions. It distinguishes a prepared examinee from a mediocre one.

Model Answer

Law of Sines: a / sin A = b / sin B = c / sin C where a, b, c are the sides of a triangle opposite to angles A, B, C respectively. Conditions for application: 1. Given two angles and one side (AAS or ASA case). 2. Given two sides and an angle opposite one of them (SSA — the ambiguous case, which may yield two, one, or no valid triangles).

Question Type

short_answer

Answer Structure

• Line 1: Write the correct Law of Sines equation [1 mark]
• Line 2: State at least one valid condition of application (AAS/ASA or SSA) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating a/sin A = b/cos B — mixing sine and cosine — loses formula mark.
• Omitting the condition and only writing the formula — loses the second mark.
• Not mentioning the ambiguous case when SSA is cited — may lose the condition mark depending on examiner.

Key Phrases To Include

• a/sin A = b/sin B = c/sin C
• AAS
• ASA
• SSA
• ambiguous case

Marks

2

Topic

Law of Cosines — Finding Angles

Difficulty

medium

Template Id

T4

Examiner Tip

When cos C is negative, the angle is obtuse (between 90° and 180°). Always check — this is a frequent source of error in board exams.

Model Answer

Given: a = 5, b = 7, c = 9 Required: Angle C By the Law of Cosines: c² = a² + b² − 2ab cos C cos C = (a² + b² − c²) / (2ab) cos C = (25 + 49 − 81) / (2 × 5 × 7) cos C = (−7) / 70 cos C = −0.1000 C = arccos(−0.1000) ∴ C ≈ 95.74°

Question Type

numerical

Answer Structure

• Line 1: State the Law of Cosines in the rearranged form for cos C [1 mark — formula/method]
• Line 2: Substitute values correctly and compute C ≈ 95.74° [1 mark — correct answer]

Scoring Breakdown

Common Mark Deductions

• Using cos C = (a² + b² + c²)/(2ab) — wrong sign — loses the formula mark.
• Not taking arccos and leaving the answer as −0.100 — loses the answer mark.
• Rounding cos C too early (e.g., to −0.1 → 84.26°) instead of correctly identifying the obtuse angle — loses the answer mark.

Key Phrases To Include

• c² = a² + b² − 2ab cos C
• cos C = (a² + b² − c²)/(2ab)
• arccos
• 95.74°

Marks

3

Topic

Trigonometric Identities — Double Angle

Difficulty

medium

Template Id

T5

Examiner Tip

Always work on one side only — the messier side. The examiner gives a full mark just for correctly applying the two double-angle identities. Never manipulate both sides simultaneously in a proof.

Model Answer

Prove: (1 − cos 2θ) / sin 2θ = tan θ Working on the Left-Hand Side (LHS): LHS = (1 − cos 2θ) / sin 2θ Step 1: Apply double-angle identities: cos 2θ = 1 − 2sin²θ → 1 − cos 2θ = 2sin²θ sin 2θ = 2 sin θ cos θ Step 2: Substitute: LHS = 2sin²θ / (2 sin θ cos θ) Step 3: Simplify: LHS = sin θ / cos θ LHS = tan θ = RHS ∴ LHS = RHS. Identity proved. ∎

Question Type

short_answer

Answer Structure

• Line 1: State which side you are working on (LHS) and write it out [0.5 mark — setup]
• Line 2: Correctly apply 1 − cos 2θ = 2sin²θ and sin 2θ = 2 sinθ cosθ [1 mark — identity substitution]
• Line 3: Simplify 2sin²θ / (2 sinθ cosθ) to sinθ / cosθ [1 mark — simplification]
• Line 4: State sinθ / cosθ = tanθ and conclude LHS = RHS [0.5 mark — conclusion statement]

Scoring Breakdown

Common Mark Deductions

• Working on both sides simultaneously (manipulating LHS and RHS in parallel) — invalid proof method, may lose all marks.
• Using cos 2θ = 2cos²θ − 1 instead of 1 − 2sin²θ — creates a dead end, likely losing simplification and conclusion marks.
• Omitting the final 'LHS = RHS, proved' statement — loses the conclusion mark.
• Arithmetic errors in simplification that change tanθ to cotθ or secθ.

Key Phrases To Include

• 1 − cos 2θ = 2sin²θ
• sin 2θ = 2 sinθ cosθ
• LHS = RHS
• tan θ = sin θ / cos θ
• Identity proved

Marks

3

Topic

Law of Sines — AAS Case and Area

Difficulty

medium

Template Id

T6

Examiner Tip

This is a classic AAS (Angle-Angle-Side) case. Always find the third angle first — it unlocks the Law of Sines application. Show the angle sum step explicitly for a guaranteed mark.

Model Answer

Given: A = 40°, C = 75°, b = 12 m Required: Side a, Area of triangle Step 1: Find angle B. B = 180° − 40° − 75° = 65° Step 2: Apply Law of Sines to find side a. a / sin A = b / sin B a / sin 40° = 12 / sin 65° a = 12 × sin 40° / sin 65° a = 12 × 0.6428 / 0.9063 a = 7714.6 / 9063 ∴ a ≈ 8.51 m Step 3: Find the area using A = ½ b·a·sin C — or use A = ½ b·c·sin A, first find c. Alternatively, use A = ½ a·b·sin C: Area = ½ × a × b × sin C Area = ½ × 8.51 × 12 × sin 75° Area = ½ × 8.51 × 12 × 0.9659 Area = ½ × 98.63 ∴ Area ≈ 49.31 m²

Question Type

numerical

Answer Structure

• Line 1: Find angle B = 180° − A − C = 65° [1 mark]
• Line 2: Apply Law of Sines correctly and compute a ≈ 8.51 m [1 mark]
• Line 3: Apply area formula A = ½ ab sin C and compute area ≈ 49.31 m² [1 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting to find angle B first and applying Law of Sines directly with only two angles given — may use wrong angle pair.
• Substituting into the area formula as ½ × b × b × sin C instead of two different sides.
• Not including units (m and m²) in the final answers — loses presentation marks.

Key Phrases To Include

• B = 180° − A − C
• a/sin A = b/sin B
• Area = ½ ab sin C
• 8.51 m
• 49.31 m²

Marks

2

Topic

Right Triangle Trigonometry — Angle of Elevation

Difficulty

easy

Template Id

T7

Examiner Tip

For angle-of-elevation problems, always use the tangent ratio when the horizontal distance and height are the two unknowns/knowns — it is the most direct approach and the examiner expects it.

Model Answer

Given: Horizontal distance = 80 m, Angle of elevation = 35° Required: Height of tower, h [Sketch: right triangle with base = 80 m, angle at base = 35°, opposite side = h] Using the tangent ratio in a right triangle: tan θ = opposite / adjacent tan 35° = h / 80 h = 80 × tan 35° h = 80 × 0.7002 ∴ h ≈ 56.02 m

Question Type

numerical

Answer Structure

• Line 1: Draw or describe the right-triangle setup [implied — no mark deduction for omitting sketch, but it organizes the solution]
• Line 2: State and apply tan 35° = h/80 [1 mark — formula and substitution]
• Line 3: Compute h = 80 × 0.7002 ≈ 56.02 m with unit [1 mark — correct answer]

Scoring Breakdown

Common Mark Deductions

• Using sin 35° = h/80 instead of tan — wrong trig function, loses both marks.
• Using cos 35° = h/80 — wrong function, loses both marks.
• Correct setup but calculation error (e.g., using tan 35° ≈ 0.57 instead of 0.70) — loses the answer mark but retains the method mark.
• Omitting the unit 'm' — may lose the final answer mark.

Key Phrases To Include

• angle of elevation
• tan 35° = h/80
• h = 80 tan 35°
• 56.02 m

Marks

3

Topic

Law of Cosines — SAS Case

Difficulty

medium

Template Id

T8

Examiner Tip

This is the classic SAS (Side-Angle-Side) problem — the most common Law of Cosines case on the board exam. Memorize c² = a² + b² − 2ab cos C. The sign is MINUS — never PLUS.

Model Answer

Given: a = 8 m, b = 6 m, C = 60° Required: (a) Side c (b) Area (a) By the Law of Cosines: c² = a² + b² − 2ab cos C c² = (8)² + (6)² − 2(8)(6) cos 60° c² = 64 + 36 − 96 × 0.5 c² = 100 − 48 c² = 52 c = √52 ∴ c ≈ 7.21 m (b) Area of the triangle: Area = ½ ab sin C Area = ½ × 8 × 6 × sin 60° Area = ½ × 48 × 0.8660 Area = 20.78 m² ∴ Area ≈ 20.78 m²

Question Type

numerical

Answer Structure

• Line 1: State Law of Cosines: c² = a² + b² − 2ab cos C [1 mark — formula]
• Line 2: Substitute and simplify to c² = 52, then c ≈ 7.21 m [1 mark — correct value of c]
• Line 3: Apply Area = ½ ab sin C = 20.78 m² [1 mark — correct area]

Scoring Breakdown

Common Mark Deductions

• Writing c² = a² + b² + 2ab cos C (wrong sign) — loses formula mark and all subsequent marks.
• Forgetting to take the square root and leaving c² = 52 as the final answer — loses the answer mark.
• Using Area = ½ base × height instead of ½ ab sin C and not identifying the correct height — likely incorrect and loses area mark.
• Omitting units in the final answer.

Key Phrases To Include

• c² = a² + b² − 2ab cos C
• cos 60° = 0.5
• c = √52 ≈ 7.21 m
• Area = ½ ab sin C
• 20.78 m²

Marks

3

Topic

Heron's Formula — Triangle Area

Difficulty

medium

Template Id

T9

Examiner Tip

's' in Heron's formula is the SEMI-perimeter — half the sum of the three sides. This is the single most common error on this topic in board exams. Write the definition of s explicitly.

Model Answer

Given: a = 5 m, b = 7 m, c = 9 m Required: Area using Heron's Formula Step 1: Compute the semi-perimeter s. s = (a + b + c) / 2 s = (5 + 7 + 9) / 2 s = 21 / 2 s = 10.5 m Step 2: Apply Heron's Formula. Area = √[s(s−a)(s−b)(s−c)] Area = √[10.5 × (10.5−5) × (10.5−7) × (10.5−9)] Area = √[10.5 × 5.5 × 3.5 × 1.5] Area = √[302.8125] ∴ Area ≈ 17.40 m²

Question Type

numerical

Answer Structure

• Line 1: Compute s = (a+b+c)/2 = 10.5 m [1 mark]
• Line 2: State and substitute into Heron's Formula √[s(s−a)(s−b)(s−c)] [1 mark]
• Line 3: Correct computation: Area ≈ 17.40 m² [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using s = a+b+c (full perimeter instead of semi-perimeter) — error propagates through all subsequent steps, loses 2 marks.
• Writing Heron's as √[(s−a)(s−b)(s−c)] without the leading 's' factor — wrong formula, loses formula mark.
• Arithmetic error in the product inside the radical — loses the final answer mark but may retain method marks.
• Not taking the square root — loses the final answer mark.

Key Phrases To Include

• s = (a+b+c)/2
• s = 10.5 m
• Area = √[s(s−a)(s−b)(s−c)]
• 17.40 m²
• semi-perimeter

Marks

5

Topic

Ambiguous Case (SSA) — Law of Sines

Difficulty

hard

Template Id

T10

Examiner Tip

The SSA ambiguous case is a premium 5-mark topic on the board exam. Examiners specifically want to see: (1) sin B calculation, (2) both B values computed, and (3) the angle-sum verification for each. Students who skip the supplement step and only give one answer will score 3/5 at best.

Model Answer

Given: a = 10 m, b = 12 m, A = 50° Required: Number of valid triangles; angle B for each case Step 1: Apply the Law of Sines to find sin B. a / sin A = b / sin B sin B = b sin A / a sin B = 12 × sin 50° / 10 sin B = 12 × 0.7660 / 10 sin B = 9.1923 / 10 sin B = 0.9192 Step 2: Find all possible values of B. Since sin B = 0.9192 < 1, solutions exist. Case 1 (acute B): B₁ = arcsin(0.9192) ≈ 66.82° Check: A + B₁ = 50° + 66.82° = 116.82° < 180° ✓ Valid Case 2 (obtuse B): B₂ = 180° − 66.82° = 113.18° Check: A + B₂ = 50° + 113.18° = 163.18° < 180° ✓ Valid Step 3: Conclusion. Both cases are valid → TWO triangles exist. Triangle 1: B₁ ≈ 66.82° Triangle 2: B₂ ≈ 113.18° ∴ Two valid triangles. B₁ ≈ 66.82°, B₂ ≈ 113.18°

Question Type

long_answer

Answer Structure

• Line 1: Set up Law of Sines: sin B = b sin A / a [1 mark]
• Line 2: Compute sin B = 0.9192 correctly [1 mark]
• Line 3: Find acute angle B₁ = 66.82° [1 mark]
• Line 4: Find obtuse angle B₂ = 180° − 66.82° = 113.18° [1 mark]
• Line 5: Verify both cases: A + B < 180° for both — confirm TWO valid triangles [1 mark]

Scoring Breakdown

Common Mark Deductions

• Finding only one value of B (the acute angle) and not checking the obtuse supplement — loses 2 marks.
• Not performing the angle-sum check to verify validity — loses the conclusion mark.
• Incorrectly computing sin B > 1 for a similar set of numbers and not recognizing it means no triangle exists.
• Writing 'one triangle' without checking whether the obtuse angle is also valid.

Key Phrases To Include

• sin B = b sin A / a
• sin B = 0.9192
• B₁ = arcsin(0.9192) ≈ 66.82°
• B₂ = 180° − B₁ = 113.18°
• ambiguous case
• two valid triangles
• A + B < 180°

Marks

5

Topic

Oblique Triangles — Bearing Applications

Difficulty

hard

Template Id

T11

Examiner Tip

Navigation/bearing problems always reward a clear sketch first. Draw the north direction, mark the bearings, and label all known angles BEFORE attempting any calculation. A correct diagram earns 1 free mark even if the calculation has an error.

Model Answer

Given: Bearing of A from ship = N 30° E Bearing of B from ship = N 80° E AB = 5 km Angle at A (ship-A-B) = 50° Required: Distance from ship (S) to lighthouse A Step 1: Determine the angle at the ship (angle ASB). Angle between bearings N 30° E and N 80° E: Angle S = 80° − 30° = 50° Step 2: Find the angle at B. Sum of angles in triangle SAB: Angle B = 180° − Angle S − Angle A Angle B = 180° − 50° − 50° = 80° Step 3: Apply the Law of Sines to find SA. AB / sin S = SA / sin B 5 / sin 50° = SA / sin 80° SA = 5 × sin 80° / sin 50° SA = 5 × 0.9848 / 0.7660 SA = 4.924 / 0.7660 ∴ SA ≈ 6.43 km

Question Type

long_answer

Answer Structure

• Line 1: Sketch triangle SAB with labelled angles and given data [1 mark — diagram/setup]
• Line 2: Compute angle at ship S = 80° − 30° = 50° [1 mark]
• Line 3: Compute angle at B = 180° − 50° − 50° = 80° [1 mark]
• Line 4: Set up and state the Law of Sines ratio correctly [1 mark]
• Line 5: Compute SA ≈ 6.43 km with unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not drawing a diagram — leads to incorrect angle identification and cascading errors.
• Confusing the angle at S and angle at A — loses angle identification marks.
• Using Law of Cosines where Law of Sines is more direct and appropriate — not wrong, but may introduce complexity and arithmetic errors.
• Omitting unit 'km' in the final answer.

Key Phrases To Include

• bearing difference = 50°
• angle sum = 180°
• AB/sin S = SA/sin B
• Law of Sines
• 6.43 km

Marks

3

Topic

Spherical Trigonometry — Spherical Excess and Area

Difficulty

hard

Template Id

T12

Examiner Tip

Spherical trigonometry appears in about 1–2 items per board exam. The spherical excess formula is simple but the −180° (not −360°) catches many students. Memorize: E = ΣAngles − 180°. Area = πR²E/180°.

Model Answer

Given: A = 95°, B = 85°, C = 100°, R = 6 m Required: (a) Spherical excess E (b) Area (a) Spherical Excess E: E = (A + B + C) − 180° E = (95° + 85° + 100°) − 180° E = 280° − 180° E = 100° (b) Area of spherical triangle: Area = (π R² × E) / 180° Area = (π × 6² × 100°) / 180° Area = (π × 36 × 100) / 180 Area = 3600π / 180 Area = 20π ∴ Area ≈ 62.83 m²

Question Type

numerical

Answer Structure

• Line 1: Compute E = (A+B+C) − 180° = 100° [1 mark]
• Line 2: State and apply Area = πR²E/180° formula [1 mark]
• Line 3: Compute Area = 20π ≈ 62.83 m² [1 mark]

Scoring Breakdown

Common Mark Deductions

• Computing E = (A+B+C) − 360° instead of −180° — wrong formula, loses all marks.
• Using the flat-triangle area formula instead of the spherical excess formula.
• Not squaring R — using R instead of R² in the formula — loses the area calculation mark.
• Giving the answer in radians or forgetting to divide by 180° when E is in degrees.

Key Phrases To Include

• E = (A+B+C) − 180°
• E = 100°
• Area = πR²E/180°
• spherical excess
• 62.83 m²

Marks

1

Topic

Sum and Difference Identities

Difficulty

easy

Template Id

T13

Examiner Tip

The cos A sin B terms cancel out by addition. This result (2 sin A cos B) is also the product-to-sum identity. Recognizing patterns like this saves time on multiple-choice items.

Model Answer

Using the sum-to-product identity: sin(A+B) = sin A cos B + cos A sin B sin(A−B) = sin A cos B − cos A sin B Adding: sin(A+B) + sin(A−B) = 2 sin A cos B ∴ sin(A+B) + sin(A−B) = 2 sin A cos B

Question Type

very_short_answer

Answer Structure

• Line 1: Expand both using sum/difference formulas (can be shown briefly) and add [1 mark — correct result 2 sin A cos B]

Scoring Breakdown

Common Mark Deductions

• Giving 2 sin A sin B — wrong, mixed up the cos B terms.
• Giving 2 cos A cos B — completely wrong identity expansion.
• Expanding only one term and giving a partial expression.

Key Phrases To Include

• sin(A+B) = sinA cosB + cosA sinB
• sin(A−B) = sinA cosB − cosA sinB
• 2 sin A cos B

Marks

5

Topic

Heron's Formula — Engineering Application

Difficulty

hard

Template Id

T14

Examiner Tip

For 5-mark problems, partial marks are awarded at each step. Even if your arithmetic is slightly off, showing the correct formula and correct setup for s and (s−a), (s−b), (s−c) guarantees at least 3 out of 5 marks. Always show all intermediate steps explicitly.

Model Answer

Given: a = 120 m, b = 150 m, c = 200 m Required: Area of triangular lot (sq. m) Step 1: Compute the semi-perimeter. s = (a + b + c) / 2 s = (120 + 150 + 200) / 2 s = 470 / 2 s = 235 m Step 2: Compute the factors (s − a), (s − b), (s − c). s − a = 235 − 120 = 115 m s − b = 235 − 150 = 85 m s − c = 235 − 200 = 35 m Step 3: Compute the product inside the radical. s(s−a)(s−b)(s−c) = 235 × 115 × 85 × 35 = 235 × 115 = 27,025 27,025 × 85 = 2,297,125 2,297,125 × 35 = 80,399,375 Step 4: Apply Heron's Formula. Area = √80,399,375 Area = √(80,399,375) ∴ Area ≈ 8,966.6 m² Note: Under RA 544, Civil Engineers are authorized to prepare and sign plans and documents relating to land development and construction — accurate area computation is a professional responsibility.

Question Type

long_answer

Answer Structure

• Line 1: Compute s = (120+150+200)/2 = 235 m [1 mark]
• Line 2: Compute all three (s−side) values correctly: 115, 85, 35 m [1 mark]
• Line 3: Compute the product s(s−a)(s−b)(s−c) = 80,399,375 [1 mark]
• Line 4: State and apply Area = √[s(s−a)(s−b)(s−c)] [1 mark — formula]
• Line 5: Correct final area ≈ 8,966.6 m² with unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using s = a+b+c = 470 m (full perimeter) — cascading error through all steps.
• Arithmetic error in one of the (s−side) values — loses that mark and potentially the product mark.
• Not showing the product computation step — loses the intermediate mark.
• Rounding intermediate values (e.g., s−a ≈ 115 rounded to 110) before the final step — significant error.

Key Phrases To Include

• s = (a+b+c)/2 = 235 m
• s−a = 115, s−b = 85, s−c = 35
• Heron's Formula
• Area = √[s(s−a)(s−b)(s−c)]
• 8,966.6 m²
• RA 544

Marks

2

Topic

Double-Angle Identities — cos 2θ

Difficulty

easy

Template Id

T15

Examiner Tip

All three forms of cos 2θ appear frequently in identity proofs and in deriving half-angle formulas. Memorize all three. The examiner specifically tests whether you know all three — listing only one or two does not earn full marks.

Model Answer

The three equivalent forms of the double-angle identity for cosine are: Form 1: cos 2θ = cos²θ − sin²θ Form 2: cos 2θ = 2cos²θ − 1 (using sin²θ = 1 − cos²θ in Form 1) Form 3: cos 2θ = 1 − 2sin²θ (using cos²θ = 1 − sin²θ in Form 1)

Question Type

short_answer

Answer Structure

• Line 1: State Form 1: cos 2θ = cos²θ − sin²θ [1 mark — must be listed]
• Line 2: State both Form 2 and Form 3 correctly [1 mark — both required for the second mark]

Scoring Breakdown

Common Mark Deductions

• Writing cos 2θ = 2sinθ cosθ — that is sin 2θ, not cos 2θ — loses both marks.
• Giving only two forms and missing one — may lose the second mark.
• Sign errors: writing cos 2θ = 2sin²θ − 1 instead of 1 − 2sin²θ — wrong form.

Key Phrases To Include

• cos²θ − sin²θ
• 2cos²θ − 1
• 1 − 2sin²θ
• double-angle identity