GELE Mathematics — Plane and Spherical TrigonometryRevision Notes
Revision notes for GELE Mathematics — Plane and Spherical Trigonometry. Short, focused, and designed for the week before exam day. Use these when you are already familiar with the chapter and need a quick refresh on the high-yield items Professional Regulation Commission (PRC) — Board of Geodetic Engineering tests.
Exam context
On the GELE 2026, the Mathematics subtest carries a "Core" weight in Professional Regulation Commission (PRC) — Board of Geodetic Engineering's pattern. Plane and Spherical Trigonometry lands at position 2nd out of 10 in the standard review order. Target score is 70% weighted average, no sub-test below 50%, and roughly a meaningful share of items come from Mathematics on a typical GELE paper.
Plane and Spherical Trigonometry - Revision Notes
Trigonometry is one of the most consistently tested topics in the PRC Civil Engineer Licensure Examination under Engineering Mathematics. It underpins surveying computations, structural force resolution, roof geometry, and geodetic calculations. This revision guide consolidates all board-critical formulas, identities, triangle-solution strategies, area formulas, and spherical trigonometry concepts into a single, exam-ready reference. Expect 3–6 items from this topic per board sitting, ranging from identity evaluation to oblique-triangle solutions and applied angle problems. Master the Law of Sines, Law of Cosines, and the ambiguous SSA case — these are perennial favorites.
Sections
Formulas
Example
If sin θ = 3/5, then cos θ = √(1 − 9/25) = 4/5 (Q1). Board items often give sin and ask for tan or sec.
Formula
sin²θ + cos²θ = 1
Variables
θ = any angle
Application
Simplify expressions, verify identities, solve for unknown trig values
Example
sec²θ − tan²θ = 1 (direct application; appears frequently in identity-type board questions)
Formula
1 + tan²θ = sec²θ | 1 + cot²θ = csc²θ
Variables
Derived from dividing sin²θ + cos²θ = 1 by cos²θ or sin²θ respectively
Application
Identity proofs and simplification of expressions involving tan and sec
Example
sin 75° = sin(45° + 30°) = sin45°cos30° + cos45°sin30° = (√6 + √2)/4 ≈ 0.9659
Formula
sin(A ± B) = sinA cosB ± cosA sinB
Variables
A, B = any angles
Application
Evaluate exact values at non-standard angles; derive double-angle identities
Example
cos 15° = cos(45° − 30°) = cos45°cos30° + sin45°sin30° = (√6 + √2)/4 ≈ 0.9659
Formula
cos(A ± B) = cosA cosB ∓ sinA sinB
Variables
Note the sign flip: cos(A+B) uses minus, cos(A−B) uses plus
Application
Exact angle evaluation; product-to-sum derivations
Example
Given sinθ = 0.6, cos θ = 0.8: sin 2θ = 2(0.6)(0.8) = 0.96; cos 2θ = 0.64 − 0.36 = 0.28
Formula
sin 2θ = 2 sinθ cosθ | cos 2θ = cos²θ − sin²θ = 1 − 2sin²θ = 2cos²θ − 1
Variables
θ = any angle; three forms of cos 2θ are all equivalent
Application
Simplify power-reduction; evaluate at double angles; used in area and integration problems
Example
tan 75° = tan(45° + 30°) = (1 + 1/√3)/(1 − 1/√3) = (√3 + 1)/(√3 − 1) = 2 + √3 ≈ 3.732
Formula
tan(A ± B) = (tanA ± tanB) / (1 ∓ tanA tanB)
Variables
A, B = any angles where the denominator is non-zero
Application
Evaluate tan at compound angles; bearing problems in surveying
Example
cos 22.5° = cos(45°/2) = √[(1 + cos45°)/2] = √[(1 + √2/2)/2] ≈ 0.9239
Formula
Half-angle: sin(θ/2) = ±√[(1 − cosθ)/2] | cos(θ/2) = ±√[(1 + cosθ)/2]
Variables
Sign depends on the quadrant of θ/2
Application
Evaluate exact values; useful in integration by trig substitution
Exam Tips
- For identity-verification board items, work one side only — never move terms across the equal sign.
- When a board item asks for sin 2θ and gives a right triangle, use sin 2θ = 2 sin θ cos θ directly from the triangle sides — no need to find the angle.
- Memorize the exact values table: sin/cos/tan at 0°, 30°, 45°, 60°, 90° — these appear constantly as given or intermediate values.
- The ASTC rule speeds up sign assignment in multi-quadrant problems: practice it until automatic.
- On the calculator, set MODE to DEGREES unless the problem explicitly states radians.
Key Points
- The six trig functions are defined from a right triangle: SOH-CAH-TOA (sin = opp/hyp, cos = adj/hyp, tan = opp/adj) and their reciprocals csc, sec, cot.
- All six functions can be expressed in terms of sin and cos alone — mastering this simplifies identity proofs.
- The three Pythagorean identities (sin²θ + cos²θ = 1; 1 + tan²θ = sec²θ; 1 + cot²θ = csc²θ) must be memorized cold.
- Sum-and-difference formulas are the parent of all multiple-angle identities — derive double-angle by setting A = B.
- Product-to-sum and sum-to-product identities appear less frequently but are useful for integration-related problems.
- Quadrant signs follow the mnemonic ASTC (All Students Take Calculus): All positive in Q1, Sin in Q2, Tan in Q3, Cos in Q4.
- Reference angle: the acute angle formed with the x-axis; used to evaluate any trig function at any angle.
- Cofunctions: sin θ = cos(90° − θ), tan θ = cot(90° − θ), sec θ = csc(90° − θ).
Definitions
Term
Pythagorean Identity
Definition
A trigonometric identity derived from the Pythagorean theorem. The fundamental form is sin²θ + cos²θ = 1, from which two secondary identities are derived by dividing through by cos²θ or sin²θ.
Importance
Foundation for simplifying all trig expressions; appears in every identity-proof board question
Term
Cofunction
Definition
A pair of trig functions that are complementary: sin and cos, tan and cot, sec and csc. sin θ = cos(90° − θ) for any angle θ.
Importance
Useful shortcut when angles sum to 90°; common in surveying and structural angle problems
Term
Reference Angle
Definition
The positive acute angle between the terminal side of any angle and the x-axis. Always between 0° and 90°.
Importance
Critical for evaluating trig functions in any quadrant, especially in board items with angles > 90°
Term
Double-Angle Formula
Definition
Identities expressing sin 2θ, cos 2θ, and tan 2θ in terms of single-angle functions. Derived by setting A = B in sum formulas.
Importance
Frequently tested directly and as intermediate steps in triangle and area problems
Section Title
1. Fundamental Trigonometric Functions and Identities
Common Mistakes
- Writing cos(A + B) = cosA + cosB — this is WRONG. Always expand using the sum formula.
- Forgetting the sign flip in cos(A + B): it uses MINUS, while cos(A − B) uses PLUS.
- Using sin 2θ = 2 sin θ (omitting the cosθ factor) — a very common careless error.
- Confusing sec with sin (sec = 1/cos, not 1/sin — that is csc).
- Not checking the quadrant when applying the Pythagorean identity to find cos θ from sin θ; the sign of cos θ depends on the quadrant.
Formulas
Example
Angle of elevation to a tower top from 50 m away (level ground) is 30°. Height = 50 × tan 30° = 50 × 0.5774 = 28.87 m
Formula
h = d · tan α
Variables
h = height of object; d = horizontal distance from observer to base; α = angle of elevation
Application
Tower height, building height, antenna problems — the most common right-triangle board item
Example
Observer moves 30 m closer; angles of elevation change from 20° to 35°. h = 30 tan 20° tan 35°/(tan 35° − tan 20°) = 30(0.364)(0.700)/(0.700 − 0.364) = 7.645/0.336 = 22.75 m
Formula
Two-observation formula: h = d · tan α · tan β / (tan β − tan α)
Variables
h = object height; d = baseline distance between observers; α = nearer angle; β = farther angle — VERIFY which is larger
Application
When only angles and the distance between two observation points are given (no direct distance to base)
Exam Tips
- Always draw a labeled diagram before writing any equation — this alone eliminates most setup errors.
- For two-observation problems, label the two horizontal distances as x and (x + d) and write two tan equations, then solve simultaneously.
- In bearing/navigation problems, convert all bearings to standard angle from East (counterclockwise) before computing vector components.
- Check: the side opposite the larger angle must be the longer side — use this as a quick sanity check.
Key Points
- Right triangle problems form the simplest class: one angle is 90°, so only two trig ratios are needed.
- Angle of elevation: measured upward from the horizontal to the line of sight toward an object above.
- Angle of depression: measured downward from the horizontal to the line of sight toward an object below.
- For two-observation problems (observer moves), set up two right triangles sharing a common height — solve simultaneously.
- Bearing problems: azimuth measured clockwise from North (0° to 360°); compass bearing uses N/S with E/W offset (e.g., N 35° E).
- Gradient or slope angle: tan θ = rise/run; used in road design and ramp problems.
- The angle of elevation from A to B equals the angle of depression from B to A (alternate interior angles).
Definitions
Term
Angle of Elevation
Definition
The angle measured upward from the horizontal line of sight to the direction of a higher object. Always positive and between 0° and 90°.
Importance
Standard setup for tower, flagpole, and building height problems — appears in nearly every board exam
Term
Angle of Depression
Definition
The angle measured downward from the horizontal to the direction of a lower object. Equal to the angle of elevation from the lower point to the higher point.
Importance
Used in navigation, surveying, and problems involving observers on elevated platforms
Term
Bearing
Definition
The direction of a line measured as an angle from a reference direction. Azimuth uses North (0°) clockwise to 360°; compass bearing uses N/S ± degrees toward E or W.
Importance
Essential for surveying traverse problems and navigation questions on the board
Section Title
2. Right Triangle Trigonometry and Angle Problems
Common Mistakes
- Confusing angle of elevation with angle of depression — draw a diagram first, always.
- Setting up d as the slant distance instead of the horizontal distance — h = d sin α, not h = d tan α, if d is the hypotenuse.
- In two-observation problems, using the wrong formula when the observer moves away instead of toward the object.
- Forgetting that bearings are from North (not East) when converting to standard mathematical angles.
Formulas
Example
Given a = 10, A = 30°, B = 45°. Find b: b = 10 × sin45° / sin30° = 10 × 0.7071 / 0.5 = 14.14
Formula
a / sinA = b / sinB = c / sinC
Variables
a, b, c = sides opposite angles A, B, C respectively
Application
SAA and ASA cases; finding sides when two angles and one side are known
Example
a = 8, b = 6, C = 60°. c² = 64 + 36 − 2(8)(6)(0.5) = 100 − 48 = 52. c = √52 = 7.21
Formula
c² = a² + b² − 2ab cosC
Variables
C = included angle between sides a and b; c = side opposite angle C
Application
SAS case (two sides + included angle); also rearranged for SSS to find any angle
Example
Sides 5, 7, 9. Find largest angle C (opposite c = 9): cosC = (25 + 49 − 81)/(2×5×7) = −7/70 = −0.1. C = arccos(−0.1) = 95.74°
Formula
cosC = (a² + b² − c²) / (2ab)
Variables
Rearrangement of Law of Cosines to find an angle given all three sides (SSS case)
Application
Finding angles in SSS triangles; checking if a triangle has an obtuse angle (cos < 0)
Example
a = 7, b = 10, A = 30°. sinB = 10 sin30°/7 = 5/7 = 0.7143. B₁ = 45.6°, B₂ = 134.4°. Check: A + B₂ = 30° + 134.4° = 164.4° < 180° ✓. Both triangles valid.
Formula
Ambiguous case check: sinB = b sinA / a; if sinB > 1 → no triangle; if sinB = 1 → right triangle; if sinB < 1 → two possible triangles (B and 180° − B, check both sum with A < 180°)
Variables
Given: a, b, A (SSA configuration)
Application
SSA problems — always perform this check to determine number of valid triangles
Exam Tips
- Memorize the case-selection rule: SAS or SSS → Law of Cosines first; anything with two angles → Law of Sines.
- After finding two angles by Law of Sines, get the third by subtraction from 180° — never apply the Law of Sines a third time (accumulates rounding error).
- In board multiple-choice, if two answer choices look like an ambiguous case pair, the problem likely has two solutions — read carefully for which one is requested.
- When sides are given as integers, the answer often involves a clean radical (√52 = 2√13) — recognize these to select quickly from MCQ choices.
- Use Law of Cosines rearranged (cosC formula) to identify obtuse triangles: negative cosine means angle > 90°.
Key Points
- Oblique triangles have NO right angle. Two tools: Law of Sines (angle-side pairs) and Law of Cosines (two sides + included angle, or all three sides).
- Case selection guide: SAA/ASA → Law of Sines; SAS/SSS → Law of Cosines; SSA → Law of Sines but check for ambiguous case.
- Law of Sines: a/sinA = b/sinB = c/sinC = 2R, where R is the circumradius.
- Law of Cosines: c² = a² + b² − 2ab cosC. Rearrange to find angle: cosC = (a² + b² − c²)/(2ab).
- Ambiguous case (SSA): given side a, side b, and angle A. Up to two triangles may exist. Always check if sin B > 1 (no solution) or if two values of B are valid.
- Sum of angles in any plane triangle = 180°. Use this to find the third angle after two are known.
- The largest angle is always opposite the longest side — use for checking solutions.
Definitions
Term
Law of Sines
Definition
States that in any triangle, the ratio of each side to the sine of its opposite angle is constant and equals twice the circumradius: a/sinA = b/sinB = c/sinC = 2R.
Importance
Primary tool for SAA, ASA, and SSA triangle cases — tested in nearly every board exam
Term
Law of Cosines
Definition
Generalizes the Pythagorean theorem to oblique triangles: c² = a² + b² − 2ab cosC. Reduces to Pythagorean theorem when C = 90°.
Importance
Only tool that works for SAS and SSS cases; also appears in vector magnitude and displacement problems
Term
Ambiguous Case (SSA)
Definition
The triangle configuration where two sides and an angle opposite one of them are given. May yield zero, one, or two valid triangles depending on the relative magnitudes of the given sides and angle.
Importance
A classic board trap — always verify the number of valid triangles and compute both if two exist
Term
Circumradius (R)
Definition
The radius of the circumscribed circle (circumcircle) of a triangle. Related to sides and angles by a/sinA = 2R.
Importance
Appears in board items asking for the circumscribed circle radius given triangle dimensions
Section Title
3. Oblique Triangles — Law of Sines and Law of Cosines
Common Mistakes
- Using Law of Sines for an SAS case — only Law of Cosines works when the given angle is between the two given sides.
- In the SSA case, checking only one value of B and missing the second valid triangle.
- Applying the Law of Cosines with the wrong angle: C must be the angle INCLUDED between sides a and b.
- Rounding intermediate values — always carry at least 4 significant figures through the calculation; round only the final answer.
- Forgetting that arcsin returns values in [−90°, 90°] — the obtuse solution B₂ = 180° − B₁ must be checked manually.
Formulas
Example
Triangle with a = 8, b = 6, C = 60°: A = ½(8)(6)sin60° = 24 × 0.8660 = 20.78 sq units
Formula
A = ½ ab sinC
Variables
a, b = two sides of the triangle; C = included angle between those two sides
Application
SAS configuration; also used after solving an oblique triangle to find its area
Example
Sides 5, 7, 9: s = (5+7+9)/2 = 10.5. A = √[10.5(5.5)(3.5)(1.5)] = √[303.1875] = 17.41 sq units
Formula
A = √[s(s−a)(s−b)(s−c)] where s = (a+b+c)/2
Variables
a, b, c = three sides; s = semi-perimeter
Application
SSS configuration — when all three sides are given but no angle is directly known
Example
Triangle with sides 3, 4, 5: s = 6, A = 6. r = 6/6 = 1 unit
Formula
r = A / s (inradius)
Variables
r = radius of inscribed circle; A = triangle area; s = semi-perimeter
Application
Board items asking for the largest circle that fits inside a triangle
Example
Triangle 3-4-5: A = 6. R = (3×4×5)/(4×6) = 60/24 = 2.5 units
Formula
R = abc / (4A) (circumradius)
Variables
R = radius of circumscribed circle; a, b, c = sides; A = triangle area
Application
Board items asking for the circumscribed circle given triangle dimensions
Exam Tips
- For a board item giving all three sides, default to Heron's formula — compute s first, then each bracket term.
- A 3-4-5 right triangle has area = 6, r = 1, R = 2.5 — memorize these benchmark values for speed.
- If a problem gives SAS, use A = ½ ab sinC first, then Law of Cosines to find c if needed — this sequence avoids extra steps.
- Area items often appear as the final part of a multi-step triangle problem — solve for the missing elements first, then apply the area formula.
Key Points
- Three area formulas cover all triangle configurations: SAS form, Heron's formula (SSS), and base-height form.
- SAS area: A = ½ab sinC — the most commonly tested area formula on the board.
- Heron's formula: A = √[s(s−a)(s−b)(s−c)] where s = (a+b+c)/2 is the SEMI-perimeter.
- The area formula A = ½ base × height applies only when the height is explicitly given or easily computed.
- Inscribed circle radius (inradius): r = Area / s; appears in board items pairing area with the inscribed circle.
- Circumscribed circle radius: R = abc / (4A); connects all three sides with the circumcircle.
- For a triangle inscribed in a circle or circumscribing a circle, area problems often combine the area formula with R or r formulas.
Definitions
Term
Semi-perimeter (s)
Definition
Half the perimeter of a triangle: s = (a + b + c)/2. Used in Heron's formula and inradius calculations.
Importance
Constantly confused with the full perimeter — always halve before using in Heron's formula
Term
Heron's Formula
Definition
An ancient formula expressing triangle area purely in terms of the three side lengths, without needing an angle. Named after Hero of Alexandria.
Importance
The go-to formula for SSS area problems; tested frequently in both standalone and multi-step board items
Term
Inradius (r)
Definition
The radius of the largest circle that fits inside a triangle, tangent to all three sides. Equal to Area divided by semi-perimeter.
Importance
Appears in geometry-linked trigonometry problems and surveying land-area items
Section Title
4. Area of Triangles
Common Mistakes
- Using s = a + b + c (full perimeter) instead of s = (a+b+c)/2 in Heron's formula — this is the single most common Heron error.
- In A = ½ ab sinC, using a non-included angle — C must be BETWEEN sides a and b.
- Forgetting to take the square root at the end of Heron's formula.
- Confusing inradius r with circumradius R — remember r = A/s and R = abc/(4A).
Formulas
Example
Given a = 50°, A = 60°, B = 45°. Find b: sin b = sin 50° × sin 45° / sin 60° = 0.7660 × 0.7071 / 0.8660 = 0.6255. b = arcsin(0.6255) = 38.73°
Formula
sin a / sinA = sin b / sinB = sin c / sinC
Variables
a, b, c = angular side lengths (arc measures); A, B, C = opposite angles — all in degrees or radians
Application
Solving spherical triangles when two angles and a side, or two sides and an angle, are known
Example
b = 60°, c = 45°, A = 80°. cos a = cos60° cos45° + sin60° sin45° cos80° = 0.5(0.7071) + 0.8660(0.7071)(0.1736) = 0.3536 + 0.1063 = 0.4599. a = arccos(0.4599) = 62.64°
Formula
cos a = cos b cos c + sin b sin c cos A
Variables
a = side opposite angle A; b, c = other two sides; A = angle between sides b and c
Application
SAS spherical triangles (two sides and included angle known); also SSS to find angles
Example
Angles 95°, 85°, 100°: E = (95 + 85 + 100) − 180 = 280 − 180 = 100°
Formula
E = (A + B + C) − 180° [spherical excess, in degrees]
Variables
A, B, C = interior angles of the spherical triangle
Application
Prerequisite for computing the area of a spherical triangle
Example
R = 6 m, E = 100°: Area = π(6)²(100)/180 = π(36)(100)/180 = 3600π/180 = 20π ≈ 62.83 m²
Formula
Area = πR²E / 180° (E in degrees) OR Area = R²E_rad (E in radians)
Variables
R = sphere radius; E = spherical excess
Application
Finding surface area of a spherical triangle on a sphere of known radius
Exam Tips
- Most board items on spherical trigonometry test only the spherical excess and area formula — master these two formulas first.
- For area problems: find E, plug into Area = πR²E/180° — that is all.
- If given a right spherical triangle (C = 90°), you can use cos c = cos a cos b directly (special case of Law of Cosines with cos C = 0).
- Angles in spherical triangles can exceed 90° — do not flag this as an error; it is geometrically valid.
Key Points
- A spherical triangle is formed by three great-circle arcs on the surface of a sphere. Sides (a, b, c) and angles (A, B, C) are all measured in angular units (degrees or radians).
- Unlike plane triangles, the sum of angles of a spherical triangle is ALWAYS greater than 180° and less than 540°.
- Spherical excess E = (A + B + C) − 180°. This is always positive and is the key to finding spherical triangle area.
- Spherical Law of Sines: sin a/sin A = sin b/sin B = sin c/sin C.
- Spherical Law of Cosines (for sides): cos a = cos b cos c + sin b sin c cos A.
- Right spherical triangles (one angle = 90°) are solved using Napier's Rules — useful shortcut for geodetic problems.
- Spherical trigonometry is used in geodesy (curved earth), astronomy, and navigation — on the board it appears primarily as area-of-spherical-triangle problems.
- Area of spherical triangle: Area = πR²E/180° where E is in degrees, or Area = R²E when E is in radians.
Definitions
Term
Great Circle
Definition
The intersection of a sphere with a plane passing through the center of the sphere. The largest circle that can be drawn on a sphere; its arcs form the sides of spherical triangles.
Importance
Conceptual foundation of spherical trigonometry; shortest path on a sphere follows a great circle
Term
Spherical Excess (E)
Definition
The amount by which the sum of angles of a spherical triangle exceeds 180°. E = A + B + C − 180°. Always positive; increases as the triangle covers more of the sphere's surface.
Importance
Directly used in the spherical triangle area formula — always compute E first
Term
Napier's Rules
Definition
A mnemonic system for solving right spherical triangles (one angle = 90°) using five circular parts. Allows quick derivation of the ten equations relating the five parts.
Importance
Efficient shortcut for geodetic board problems involving right spherical triangles
Section Title
5. Spherical Trigonometry
Common Mistakes
- Treating spherical triangles like plane triangles — the angle sum is NOT 180°; it is 180° + E.
- Using degrees in the area formula but forgetting to convert E to radians when the formula uses E_rad — always check which form is being applied.
- Applying the plane Law of Cosines (c² = a² + b² − 2ab cosC) to spherical triangles — spherical Law of Cosines uses cos a = cos b cos c + sin b sin c cos A (all cosines, no squares).
- Confusing angular side lengths with linear arc lengths — convert: arc length = Rθ (in radians).
Connections
- SURVEYING: Traverse computation uses bearing-to-angle conversion and the Law of Sines/Cosines for triangle closure; area by DMD or coordinate method uses trig functions directly.
- STRUCTURAL ENGINEERING: Force resolution into components uses sinθ and cosθ; roof truss geometry applies Law of Sines for member-length calculation.
- GEODESY / ENGINEERING SURVEYS: Spherical trigonometry is used when the earth's curvature is significant (long baselines > 10 km); geodetic coordinates require spherical excess corrections.
- ALGEBRA AND ANALYTIC GEOMETRY: The distance formula and dot product of vectors use the Law of Cosines concept; the cross product magnitude uses sinθ.
- CALCULUS: Trig identities (especially double-angle and half-angle) are prerequisite to trig substitution in integration — board items on areas under curves may require these.
- PHYSICS / ENGINEERING MECHANICS: Resolution of inclined forces, moments, and projectile motion all rely on sinθ and cosθ decomposition.
- RA 544 (Civil Engineering Law): Professional practice of surveying and geodesy, which uses trigonometry as its computational backbone, is regulated under this law.
Exam Strategy
In the PRC CE Board Exam, Plane and Spherical Trigonometry items typically appear in the Engineering Mathematics afternoon session. Strategy: (1) Identify the given information first and classify the triangle type (right, SAA, ASA, SAS, SSS, SSA) — this immediately tells you which formula to use. (2) For identity problems, work on the more complex side and transform it toward the simpler side using Pythagorean identities and algebraic manipulation — never move terms across the equals sign. (3) For applied angle problems (elevation, depression, bearings), draw the diagram before writing equations — this eliminates setup errors. (4) In multiple-choice, estimate the answer using approximate trig values (sin 30° = 0.5, cos 60° = 0.5, sin 45° ≈ 0.707) to eliminate unreasonable choices quickly. (5) For Heron's formula, compute s first and write it down explicitly — most errors occur by using the full perimeter. (6) For spherical triangle area, always compute E = A + B + C − 180° first, then apply Area = πR²E/180°. (7) Manage time: straightforward trig items should take 1.5–2 minutes each; multi-step triangle-solution items may take 3–4 minutes — skip and return if stuck. (8) Always verify your answer: use the angle-sum check (A+B+C = 180° for plane, >180° for spherical) and the side-angle ordering rule (larger side opposite larger angle).
Quick Review Questions
A triangle has sides a = 5, b = 7, c = 9. Using Heron's formula, find the area.
s = (5+7+9)/2 = 10.5. A = √[10.5(10.5−5)(10.5−7)(10.5−9)] = √[10.5 × 5.5 × 3.5 × 1.5] = √[303.1875] = 17.41 sq units. Note: s is the SEMI-perimeter, not the full perimeter.
In a triangle, a = 10, A = 30°, B = 45°. Find side b using the Law of Sines.
By Law of Sines: b/sinB = a/sinA → b = a sinB/sinA = 10 × sin45°/sin30° = 10 × 0.7071/0.5 = 14.14. First find C = 180° − 30° − 45° = 105° to confirm the triangle is valid.
A triangle has a = 8, b = 6, with included angle C = 60°. Find side c and the area.
Law of Cosines: c² = 64 + 36 − 2(8)(6)cos60° = 100 − 48 = 52; c = 7.21. Area = ½ab sinC = ½(8)(6)sin60° = 24 × 0.8660 = 20.78 sq units.
The angle of elevation to the top of a flagpole from a point 40 m away (level ground) is 35°. How tall is the flagpole?
h = d × tan α = 40 × tan 35° = 40 × 0.7002 = 28.01 m. Always use the horizontal distance (not slant) with tangent.
A spherical triangle has angles A = 95°, B = 85°, C = 100° on a sphere of radius 6 m. Find the area of the spherical triangle.
Spherical excess: E = (95 + 85 + 100) − 180 = 100°. Area = πR²E/180° = π(36)(100)/180 = 3600π/180 = 20π ≈ 62.83 m².
Verify the identity: (1 − sin²θ)/cosθ = cosθ
Left side: (1 − sin²θ)/cosθ = cos²θ/cosθ = cosθ. Used the Pythagorean identity sin²θ + cos²θ = 1, so 1 − sin²θ = cos²θ. Only one side was manipulated — correct identity-proof technique.
In an SSA case: a = 7, b = 10, A = 30°. How many valid triangles exist? Find all valid angles B.
sinB = b sinA/a = 10 × 0.5/7 = 0.7143. B₁ = arcsin(0.7143) = 45.58°. B₂ = 180° − 45.58° = 134.42°. Check: A + B₂ = 30° + 134.42° = 164.42° < 180° ✓. Both are valid — ambiguous case with two triangles.
Find the exact value of sin 75°.
sin 75° = sin(45° + 30°) = sin45°cos30° + cos45°sin30° = (√2/2)(√3/2) + (√2/2)(1/2) = √6/4 + √2/4 = (√6 + √2)/4. This tests the sum formula directly.
Find all three angles of a triangle with sides a = 5, b = 7, c = 9 (use Law of Cosines).
cosC = (25+49−81)/70 = −7/70 = −0.1000 → C = 95.74°. cosB = (25+81−49)/90 = 57/90 = 0.6333 → B = 50.70°. A = 180° − 95.74° − 50.70° = 33.56°. Check: 33.56 + 50.70 + 95.74 = 180° ✓
A 3-4-5 right triangle: find the inradius r and circumradius R.
Area = ½(3)(4) = 6. s = (3+4+5)/2 = 6. r = Area/s = 6/6 = 1. R = abc/(4A) = (3×4×5)/(4×6) = 60/24 = 2.5. Also R = hypotenuse/2 = 5/2 = 2.5 (for right triangles). Memorize these benchmark values.
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