Plane and Spherical Trigonometry - Detailed Explanation
Trigonometry is one of the most heavily tested topics in the PRC Civil Engineer Licensure Examination under Engineering Mathematics. It forms the mathematical backbone of surveying, structural analysis, force resolution, and geodesy. This chapter covers plane trigonometric functions and identities, oblique triangle solution methods (Law of Sines and Law of Cosines), area formulas, and an introduction to spherical trigonometry as applied in geodetic problems. Mastery of these concepts is non-negotiable: expect 5–10 items per board examination directly or indirectly involving trigonometry. Study each concept with its worked examples, understand the decision logic for choosing the right formula, and drill the common pitfalls to maximize your score.
Trigonometric Functions of a Right Triangle
Examples
A right triangle has an angle of 35° and the side adjacent to that angle measures 12 m. Find the opposite side and the hypotenuse.
Solution
Step 1 — Find the opposite side using tangent:
tan 35° = opp / adj
opp = adj × tan 35° = 12 × 0.7002 = 8.40 m
Step 2 — Find the hypotenuse using cosine:
cos 35° = adj / hyp
hyp = adj / cos 35° = 12 / 0.8192 = 14.65 m
Step 3 — Verify using Pythagorean theorem:
hyp² = 12² + 8.40² = 144 + 70.56 = 214.56
hyp = √214.56 = 14.65 m ✓
Always identify which side is given, then select the trig function that relates the given side to the unknown. Use the Pythagorean theorem as a verification tool whenever possible.
Given sin θ = 3/5 and θ is in Quadrant II, find cos θ, tan θ, and csc θ.
Solution
Step 1 — Find the third side using Pythagorean identity:
sin²θ + cos²θ = 1
(3/5)² + cos²θ = 1
9/25 + cos²θ = 1
cos²θ = 16/25
cos θ = ±4/5
Step 2 — Apply ASTC: Q2 → cosine is negative:
cos θ = −4/5
Step 3 — Find tan θ:
tan θ = sin θ / cos θ = (3/5) / (−4/5) = −3/4
Step 4 — Find csc θ:
csc θ = 1 / sin θ = 5/3
The Pythagorean identity always gives two possible values for the unknown function. The quadrant information eliminates the ambiguity. This is a classic board exam question pattern.
The six trigonometric functions are defined with respect to an acute angle θ in a right triangle. Label the sides relative to θ as: opposite (opp), adjacent (adj), and hypotenuse (hyp). The primary functions are:
- sin θ = opp / hyp
- cos θ = adj / hyp
- tan θ = opp / adj
The reciprocal functions are:
- csc θ = hyp / opp = 1 / sin θ
- sec θ = hyp / adj = 1 / cos θ
- cot θ = adj / opp = 1 / tan θ
Mnemonic: SOH-CAH-TOA (Sine = Opposite over Hypotenuse, Cosine = Adjacent over Hypotenuse, Tangent = Opposite over Adjacent).
Signs per quadrant (ASTC rule — All Students Take Calculus):
- Q1: All positive
- Q2: Sine positive (and csc)
- Q3: Tangent positive (and cot)
- Q4: Cosine positive (and sec)
Reference angles: The reference angle is the acute angle formed with the x-axis. For any angle θ in Q2: ref = 180° − θ; Q3: ref = θ − 180°; Q4: ref = 360° − θ.
Special angle values you must memorize:
- sin 0° = 0, cos 0° = 1, tan 0° = 0
- sin 30° = 0.5, cos 30° = √3/2 ≈ 0.866, tan 30° = 1/√3 ≈ 0.577
- sin 45° = √2/2 ≈ 0.707, cos 45° = √2/2, tan 45° = 1
- sin 60° = √3/2 ≈ 0.866, cos 60° = 0.5, tan 60° = √3 ≈ 1.732
- sin 90° = 1, cos 90° = 0, tan 90° = undefined
Applications
- Surveying: computing horizontal and vertical distances from slope distances and vertical angles.
- Structural analysis: resolving forces into horizontal and vertical components.
- Highway engineering: computing superelevation and grade calculations.
- Architectural drawing: computing rafter lengths and roof pitch.
- Setting out: computing coordinates from bearings and distances.
Common misconceptions
- Confusing csc with cos — csc is the reciprocal of SINE, not cosine.
- Forgetting to check the quadrant when finding trig values from identities — always two answers before applying ASTC.
- Applying SOH-CAH-TOA to oblique (non-right) triangles — it ONLY works for right triangles.
- Using radian mode on calculator when the problem uses degrees — always confirm calculator mode.
- Confusing angle of elevation (measured upward from horizontal) with angle of depression (measured downward from horizontal).
Related concepts
- Pythagorean Theorem
- Trigonometric Identities
- Law of Sines
- Law of Cosines
- Inverse Trigonometric Functions
Common exam questions
Example
If cos θ = −5/13 and sin θ > 0, find all six trig functions. Answer: Q2 → sin = 12/13, tan = −12/5, csc = 13/12, sec = −13/5, cot = −5/12.
Approach
Use sin²θ + cos²θ = 1 to find the missing primary function, apply ASTC for sign, then derive the rest from definitions and reciprocals.
Example
A 40-m tall building casts a shadow of 55 m on level ground. Find the angle of elevation of the sun. Solution: tan θ = 40/55 → θ = arctan(0.7273) = 36.03°.
Approach
Draw a right triangle. Identify the known side and the required side. Select the appropriate trig ratio. Set calculator to degree mode.
Example
A ramp makes a 15° angle with the horizontal. If the ramp is 20 m long, find its vertical rise. Rise = 20 sin 15° = 20 × 0.2588 = 5.18 m.
Approach
Direct application of sin or cos. opp = hyp × sin θ; adj = hyp × cos θ.
Key points to remember
- SOH-CAH-TOA is the foundation — know it instantly without thinking.
- ASTC rule determines the sign of any trig function in any quadrant.
- Reference angle always measured from the x-axis, always positive and acute.
- Calculator must be in DEGREE mode unless problem explicitly states radians.
- Reciprocal functions: csc = 1/sin, sec = 1/cos, cot = 1/tan — never confuse csc with cos.
- Special angles (30°, 45°, 60°) frequently appear without a calculator allowed — memorize exact values.
- For complementary angles: sin θ = cos(90° − θ), tan θ = cot(90° − θ), sec θ = csc(90° − θ).
Fundamental Trigonometric Identities
Examples
Simplify: (sin²θ − 1) / cos θ
Solution
Step 1 — Use Pythagorean identity: sin²θ − 1 = −cos²θ
Step 2 — Substitute:
(sin²θ − 1) / cos θ = −cos²θ / cos θ
Step 3 — Simplify:
= −cos θ
Final Answer: −cos θ
The key move is recognizing that sin²θ − 1 = −(1 − sin²θ) = −cos²θ. This type of simplification appears frequently in board exams.
Find the exact value of sin 75°.
Solution
Step 1 — Express 75° as a sum of known angles:
75° = 45° + 30°
Step 2 — Apply sum formula:
sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30°
Step 3 — Substitute known values:
= (√2/2)(√3/2) + (√2/2)(1/2)
= √6/4 + √2/4
= (√6 + √2)/4
Decimal check: (2.449 + 1.414)/4 = 3.863/4 = 0.9659
Verify: sin 75° = 0.9659 ✓
Sum-and-difference formulas let you compute exact values for angles that are not in the standard table. This technique appears regularly in Philippine board exams.
If sin θ = 4/5 and θ is in Q1, find sin 2θ, cos 2θ, and tan 2θ.
Solution
Step 1 — Find cos θ using Pythagorean identity:
cos θ = √(1 − 16/25) = √(9/25) = 3/5 (positive, Q1)
Step 2 — sin 2θ:
sin 2θ = 2 sin θ cos θ = 2(4/5)(3/5) = 24/25
Step 3 — cos 2θ:
cos 2θ = cos²θ − sin²θ = 9/25 − 16/25 = −7/25
Step 4 — tan 2θ:
tan 2θ = sin 2θ / cos 2θ = (24/25)/(−7/25) = −24/7
Note: 2θ is in Q2 (sin positive, cos negative, tan negative) — consistent with results.
Double-angle problems require first finding all primary trig functions, then applying the double-angle formulas. Always verify the quadrant of 2θ to confirm signs.
Trigonometric identities are equations that are true for all valid values of the variable. They are essential tools for simplifying expressions and solving equations in board exams.
Pythagorean Identities (Most Important):
- sin²θ + cos²θ = 1
- 1 + tan²θ = sec²θ
- 1 + cot²θ = csc²θ
Derivation tip: Divide the first identity by cos²θ to get the second; divide by sin²θ to get the third.
Reciprocal Identities:
- csc θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ
Quotient Identities:
- tan θ = sin θ / cos θ
- cot θ = cos θ / sin θ
Even-Odd Identities:
- sin(−θ) = −sin θ (odd function)
- cos(−θ) = cos θ (even function)
- tan(−θ) = −tan θ (odd function)
Sum and Difference Formulas:
- sin(A ± B) = sin A cos B ± cos A sin B
- cos(A ± B) = cos A cos B ∓ sin A sin B
- tan(A ± B) = (tan A ± tan B) / (1 ∓ tan A tan B)
Double Angle Formulas:
- sin 2θ = 2 sin θ cos θ
- cos 2θ = cos²θ − sin²θ = 1 − 2sin²θ = 2cos²θ − 1
- tan 2θ = 2tan θ / (1 − tan²θ)
Half Angle Formulas:
- sin(θ/2) = ±√[(1 − cos θ)/2]
- cos(θ/2) = ±√[(1 + cos θ)/2]
- tan(θ/2) = (1 − cos θ)/sin θ = sin θ/(1 + cos θ)
Product-to-Sum and Sum-to-Product:
- sin A + sin B = 2 sin[(A+B)/2] cos[(A−B)/2]
- cos A + cos B = 2 cos[(A+B)/2] cos[(A−B)/2]
- sin A cos B = ½[sin(A+B) + sin(A−B)]
Applications
- Simplifying complex trigonometric expressions in structural mechanics.
- Exact value computation for angle problems in surveying calculations.
- Proving trigonometric equations in mathematical analysis.
- Computing resultant forces in vector mechanics using sum/difference angles.
- Signal and wave problems in electrical engineering (product-to-sum formulas).
Common misconceptions
- sin(A + B) ≠ sin A + sin B — the sum formula MUST be applied. This is a very common board exam trap.
- cos 2θ has THREE equivalent forms — using the wrong form can complicate the problem unnecessarily. Choose based on what's given.
- (sin θ)² is written as sin²θ, NOT sin(θ²). The exponent applies to the function value, not the angle.
- √(sin²θ) = |sin θ|, not simply sin θ — the absolute value matters when the quadrant is unspecified.
- When proving identities, you cannot cross-multiply across the equals sign — work only one side at a time.
Related concepts
- Trigonometric Functions
- Solving Trigonometric Equations
- Oblique Triangle Solutions
- Complex Numbers (Euler's Formula)
- Fourier Analysis
Common exam questions
Example
Simplify: (1 − cos 2θ) / sin 2θ. Apply identities: 1 − cos 2θ = 2sin²θ, sin 2θ = 2sinθcosθ. Result = 2sin²θ / (2sinθcosθ) = sinθ/cosθ = tan θ.
Approach
Apply Pythagorean identities to simplify; factor if possible; look for sin²+cos²=1 patterns. Board exams often disguise sec²θ − tan²θ = 1.
Example
Find cos 15°. cos 15° = cos(45°−30°) = cos45°cos30° + sin45°sin30° = (√6+√2)/4 ≈ 0.9659.
Approach
Express the given angle as sum or difference of 30°, 45°, or 60°. Apply the appropriate formula.
Example
Prove: sec²θ − 1 = tan²θ. LHS: sec²θ − 1 = (1/cos²θ) − 1 = (1 − cos²θ)/cos²θ = sin²θ/cos²θ = tan²θ = RHS. ✓
Approach
Start from the more complex side. Rewrite everything in terms of sin and cos. Apply Pythagorean identities. Never move terms across the equals sign.
Key points to remember
- sin²θ + cos²θ = 1 is the master identity — all Pythagorean identities derive from it.
- For double-angle cos 2θ, you have THREE equivalent forms — choose the one that makes the problem simplest.
- Sum/difference formulas are frequently used to find exact values of non-standard angles (e.g., sin 75° = sin(45°+30°)).
- Memorize the sign pattern for sum/difference: sin has same sign as ±, cos has OPPOSITE sign.
- Half-angle sign (±) depends on the quadrant of θ/2.
- tan(A+B) denominator has MINUS when signs are both plus; tan(A−B) has PLUS when signs alternate.
- Product-to-sum formulas appear in signal processing and wave analysis problems.
Oblique Triangle Solutions — Law of Sines
Examples
In triangle ABC, A = 40°, B = 75°, and a = 20 m. Find b, c, and the area.
Solution
Step 1 — Find angle C:
C = 180° − 40° − 75° = 65°
Step 2 — Apply Law of Sines to find b:
b / sin B = a / sin A
b = a × sin B / sin A = 20 × sin 75° / sin 40°
b = 20 × 0.9659 / 0.6428 = 30.08 m
Step 3 — Find c:
c = a × sin C / sin A = 20 × sin 65° / sin 40°
c = 20 × 0.9063 / 0.6428 = 28.22 m
Step 4 — Find area:
Area = (1/2) × b × c × sin A = (1/2)(30.08)(28.22) × sin 40°
Area = (1/2)(30.08)(28.22)(0.6428)
Area = 272.86 m²
Alternatively: Area = (1/2) × a × b × sin C = (1/2)(20)(30.08) × sin 65° = 272.86 m² ✓
AAS case: always find the third angle first, then use the Law of Sines ratio. Use any pair of sides and their included angle to compute area.
Given a = 10 m, b = 14 m, A = 30°. Determine how many triangles exist and solve them.
Solution
Step 1 — Check if A < 90°: Yes (30° < 90°).
Step 2 — Compute h = b sin A = 14 × sin 30° = 14 × 0.5 = 7 m
Step 3 — Compare a with h and b:
h = 7 m, a = 10 m, b = 14 m
Since h < a < b (7 < 10 < 14), TWO triangles exist.
Step 4 — Find B₁:
sin B = b sin A / a = 14 × 0.5 / 10 = 0.7
B₁ = arcsin(0.7) = 44.43°
Step 5 — First triangle:
C₁ = 180° − 30° − 44.43° = 105.57°
c₁ = a sin C₁ / sin A = 10 × sin(105.57°) / sin(30°) = 10 × 0.9636 / 0.5 = 19.27 m
Step 6 — Second triangle:
B₂ = 180° − 44.43° = 135.57°
C₂ = 180° − 30° − 135.57° = 14.43°
c₂ = a sin C₂ / sin A = 10 × sin(14.43°) / 0.5 = 10 × 0.2492 / 0.5 = 4.98 m
The ambiguous case requires checking the height h = b sin A first. When h < a < b, you ALWAYS get two valid triangles. Both must be solved completely. Board exams may ask for the area of either or both triangles.
An oblique triangle is any triangle that is NOT a right triangle. Two major laws govern oblique triangles: the Law of Sines and the Law of Cosines.
Law of Sines:
a / sin A = b / sin B = c / sin C = 2R
where a, b, c are the sides opposite to angles A, B, C respectively, and R is the circumradius of the triangle.
The Law of Sines is used when you know:
- Case 1 (AAS): Two angles and one side (any side)
- Case 2 (ASA): Two angles and the included side
- Case 3 (SSA): Two sides and an angle opposite one of them — WARNING: Ambiguous Case!
Solving AAS / ASA:
- Find the third angle: C = 180° − A − B
- Apply the ratio: b = a sin B / sin A
- Find the remaining side similarly
The Ambiguous Case (SSA):
Given sides a, b and angle A (where a is opposite A and b is adjacent):
- If A ≥ 90°: Only one triangle if a > b; no triangle if a ≤ b.
If A < 90°: Compute h = b sin A
- If a < h: No triangle (a too short to reach the base)
- If a = h: Exactly one right triangle
- If h < a < b: Two triangles (ambiguous case — two valid values of angle B)
- If a ≥ b: Exactly one triangle
For the two-triangle case, the two values of B are:
B₁ = arcsin(b sin A / a) and B₂ = 180° − B₁
Check that A + B₂ < 180° for the second triangle to be valid.
Applications
- Surveying: computing unknown distances and angles in triangulation networks.
- Navigation: determining ship position using two known bearings from two known points.
- Construction layout: setting out triangular lots when only partial measurements are available.
- Civil engineering: computing span lengths and cable tensions in cable-stayed structures.
- Geodesy: triangulation for horizontal control point determination.
Common misconceptions
- SSA is not always ambiguous — it depends on whether h < a < b. Many students automatically assume two triangles for all SSA cases.
- Law of Sines cannot be applied without a complete angle-side pair. If you only have two sides and the included angle, you need Law of Cosines first.
- The second triangle in the ambiguous case uses B₂ = 180° − B₁, and you must still verify A + B₂ < 180°.
- Area = (1/2) a b sin C requires the INCLUDED angle C between sides a and b, not any angle.
- Never divide by sin 0° = 0 — this happens if an angle is 0° or 180°, which is impossible in a real triangle.
Related concepts
- Law of Cosines
- Triangle Area Formulas
- Circumscribed and Inscribed Circles
- Surveying Triangulation
- Vector Resolution
Common exam questions
Example
Triangle: A = 55°, C = 80°, b = 15 m. Find a. First: B = 180° − 55° − 80° = 45°. Then: a = b sin A / sin B = 15 × sin55°/sin45° = 15 × 0.8192/0.7071 = 17.38 m.
Approach
Find third angle first (180° − A − B), apply sines ratio to find sides, use Area = (1/2)ab sin C.
Example
a = 8, b = 10, A = 35°. h = 10 sin 35° = 5.74. Since h(5.74) < a(8) < b(10), TWO triangles. sin B = 10 sin35°/8 = 0.7182, B₁ = 45.84°, B₂ = 134.16°. Both valid since A + B₂ = 169.16° < 180°.
Approach
Compute h = b sin A. Compare a vs. h and b. State how many triangles. Solve for both if two exist.
Example
In a triangle with a = 12 m and A = 40°, R = 12 / (2 sin 40°) = 12 / (2 × 0.6428) = 9.33 m.
Approach
Use R = a / (2 sin A). Straightforward once the triangle is solved.
Key points to remember
- Law of Sines requires an ANGLE-OPPOSITE SIDE pair to set up the ratio.
- AAS and ASA give unique triangles; SSA may give 0, 1, or 2 valid triangles.
- The third angle is always found by: third angle = 180° − (sum of known angles).
- The ambiguous case (SSA) is one of the most tested board exam pitfalls — always check the number of valid triangles.
- The ratio a / sin A = 2R links the law of sines to the circumscribed circle radius R.
- Always verify: sum of all three angles must equal exactly 180°.
- Use law of sines when you have a complete angle-side pair (one angle and its opposite side both known).
Oblique Triangle Solutions — Law of Cosines
Examples
A triangle has sides a = 8 m, b = 6 m, with included angle C = 60°. Find side c, angles A and B, and the area. (Classic board exam problem)
Solution
Step 1 — Find c using Law of Cosines:
c² = a² + b² − 2ab cos C
c² = 8² + 6² − 2(8)(6) cos 60°
c² = 64 + 36 − 96(0.5)
c² = 100 − 48 = 52
c = √52 = 7.21 m
Step 2 — Find angle A using Law of Cosines:
cos A = (b² + c² − a²) / (2bc)
cos A = (36 + 52 − 64) / (2 × 6 × 7.21)
cos A = 24 / 86.52 = 0.2774
A = arccos(0.2774) = 73.90°
Step 3 — Find angle B:
B = 180° − 60° − 73.90° = 46.10°
Verify: cos B = (64 + 52 − 36) / (2 × 8 × 7.21) = 80/115.36 = 0.6935 → B = 46.10° ✓
Step 4 — Find area:
Area = (1/2) × a × b × sin C = (1/2)(8)(6) sin 60°
Area = 24 × 0.8660 = 20.78 m²
This is the most common board exam application of Law of Cosines (SAS case). The formula structure c² = a² + b² − 2ab cos C is the one to memorize. Note how all three angles now sum to 180°: 60° + 73.90° + 46.10° = 180° ✓.
A triangle has sides 5 m, 7 m, and 9 m. Find all three angles.
Solution
Step 1 — Identify the largest side: c = 9 m (opposite angle C). Find C first.
cos C = (a² + b² − c²) / (2ab)
cos C = (25 + 49 − 81) / (2 × 5 × 7)
cos C = −7 / 70 = −0.1000
C = arccos(−0.1000) = 95.74° (obtuse — correctly handled by Law of Cosines)
Step 2 — Find angle A (or B) using Law of Cosines:
cos A = (b² + c² − a²) / (2bc)
cos A = (49 + 81 − 25) / (2 × 7 × 9)
cos A = 105 / 126 = 0.8333
A = arccos(0.8333) = 33.56°
Step 3 — Find B by subtraction:
B = 180° − 95.74° − 33.56° = 50.70°
Verification: 33.56° + 50.70° + 95.74° = 180.00° ✓
The SSS case always starts with the largest angle. Here, C = 95.74° is obtuse, which would be missed if you used arcsin (arcsin returns acute angles only). Always use arccos for Law of Cosines angle calculations.
The Law of Cosines is used when the Law of Sines cannot be directly applied — specifically when you have:
- Case SAS: Two sides and the included angle
- Case SSS: All three sides known (to find angles)
Law of Cosines (Three forms):
a² = b² + c² − 2bc cos A
b² = a² + c² − 2ac cos B
c² = a² + b² − 2ab cos C
Solving for angles from three known sides:
cos A = (b² + c² − a²) / (2bc)
cos B = (a² + c² − b²) / (2ac)
cos C = (a² + b² − c²) / (2ab)
Important: If the computed value of cos A is negative, then A > 90° (obtuse angle). This is perfectly valid.
SAS Solution Procedure:
- Use Law of Cosines to find the unknown side.
- Use Law of Sines (or Cosines) to find one of the unknown angles.
- Find the third angle by subtraction from 180°.
SSS Solution Procedure:
- Use Law of Cosines to find the largest angle first (opposite the longest side).
- Use Law of Sines or Cosines to find another angle.
- Find the third by subtraction.
Why find the largest angle first in SSS? Because the largest angle might be obtuse. Using the Law of Sines for an obtuse angle introduces ambiguity (arcsin gives the acute supplement). By finding the largest angle with the Law of Cosines, which gives the correct obtuse angle directly, you eliminate the ambiguity for the remaining (necessarily acute) angles.
Applications
- Structural engineering: computing member lengths in trusses given joint coordinates.
- Route surveying: computing traverse closure when all distances are measured.
- Hydrology: computing distances in drainage basin geometry.
- Geodetics: computing baselines in triangulation surveys.
- Force analysis: resolving resultant of two non-perpendicular forces.
Common misconceptions
- c² = a² + b² − 2ab cos C — the included angle must be C (between a and b). Using the wrong angle gives a completely wrong answer.
- In SSS, using Law of Sines to find the first angle may give the wrong result if that angle is obtuse (arcsin only returns acute angles).
- The Law of Cosines applies to ALL triangles, including right triangles — it simply reduces to Pythagorean theorem when the angle is 90°.
- Forgetting the negative sign in front of 2ab cos C is a very common computation error. Double-check the formula structure.
- Heron's formula uses s = (a+b+c)/2 (SEMI-perimeter), not the full perimeter.
Related concepts
- Law of Sines
- Area of a Triangle
- Pythagorean Theorem
- Vector Dot Product (related geometric interpretation)
- Coordinate Geometry
Common exam questions
Example
Two sides of a triangle are 10 m and 15 m, included angle 120°. Find the third side. c² = 100 + 225 − 2(10)(15)cos120° = 325 − 300(−0.5) = 325 + 150 = 475. c = 21.79 m.
Approach
Direct application: c² = a² + b² − 2ab cos C. Substitute and solve. Check: result should be between |a−b| and (a+b).
Example
Sides 6, 8, 11. Largest is 11. cos C = (36+64−121)/(2×6×8) = −21/96 = −0.2188. C = arccos(−0.2188) = 102.62°.
Approach
Identify the longest side. Use cos A = (b²+c²−a²)/(2bc) where a is the longest side. Result for cos might be negative (obtuse angle).
Example
Sides 5, 7, 9. s = 21/2 = 10.5. Area = √[10.5×5.5×3.5×1.5] = √[302.8125] = 17.40 m².
Approach
Compute s = (a+b+c)/2. Then Area = √[s(s−a)(s−b)(s−c)]. Cross-check with (1/2)ab sin C if angle is known.
Key points to remember
- Law of Cosines = generalization of Pythagorean theorem. When C = 90°, 2ab cos 90° = 0, reducing to c² = a² + b².
- The angle in cos C must be the angle BETWEEN sides a and b (included angle).
- Negative value of cos C means C is obtuse — the formula handles this correctly.
- SAS gives unique triangle; SSS gives unique triangle; both use Law of Cosines.
- In SSS, always find the LARGEST angle first using Law of Cosines to avoid the arcsin ambiguity.
- After finding one angle via Law of Cosines, you may use Law of Sines for the remaining angles if they are not the largest.
- Always verify: A + B + C = 180° at the end.
Area Formulas for Triangles
Examples
A triangular lot has sides of 50 m, 70 m, and 90 m. Compute the area using Heron's formula and verify with another method.
Solution
Step 1 — Compute semi-perimeter:
s = (50 + 70 + 90) / 2 = 210 / 2 = 105 m
Step 2 — Compute s−a, s−b, s−c:
s − 50 = 55 m
s − 70 = 35 m
s − 90 = 15 m
Step 3 — Apply Heron's formula:
Area = √[105 × 55 × 35 × 15]
= √[105 × 55 × 35 × 15]
= √[3,031,875]
= 1,741.2 m²
Step 4 — Verification using Area = (1/2)ab sin C:
First find angle C (opposite 90 m side):
cos C = (50² + 70² − 90²) / (2 × 50 × 70)
= (2500 + 4900 − 8100) / 7000
= −700 / 7000 = −0.1000
C = 95.74°
Area = (1/2)(50)(70) sin 95.74°
= 1750 × 0.9950 = 1741.2 m² ✓
Heron's formula works directly from the three sides without needing any angles. The verification via (1/2)ab sin C (after computing C via Law of Cosines) confirms the result. Both methods yield 1,741.2 m².
A triangle has sides a = 12 m and b = 15 m with included angle C = 50°. Find the area, circumradius R, and inradius r.
Solution
Step 1 — Area:
Area = (1/2) × 12 × 15 × sin 50° = (1/2)(12)(15)(0.7660) = 68.94 m²
Step 2 — Find side c using Law of Cosines:
c² = 144 + 225 − 2(12)(15)(0.6428)
= 369 − 231.4 = 137.6
c = 11.73 m
Step 3 — Circumradius R:
First find angle A using Law of Sines:
sin A = a sin C / c = 12 × 0.7660 / 11.73 = 0.7839
A = 51.64°
R = a / (2 sin A) = 12 / (2 × 0.7839) = 7.65 m
Step 4 — Inradius r:
s = (12 + 15 + 11.73) / 2 = 38.73 / 2 = 19.365 m
r = Area / s = 68.94 / 19.365 = 3.56 m
This problem integrates multiple concepts: area by SAS, law of cosines, law of sines, circumradius from law of sines, and inradius from area/semi-perimeter. Board exams often chain these calculations.
Computing the area of a triangle in various given conditions is a staple of board examinations. The appropriate formula depends on what information is given.
Formula 1 — Base and Height (most basic):
Area = (1/2) × base × height = (1/2) bh
Formula 2 — Two Sides and Included Angle (SAS):
Area = (1/2) ab sin C
where C is the angle included between sides a and b.
This is the most frequently used formula in board exams for oblique triangles.
Formula 3 — Heron's Formula (SSS):
Area = √[s(s−a)(s−b)(s−c)]
where s = (a+b+c)/2 is the semi-perimeter.
Formula 4 — Circumradius Formula:
Area = abc / (4R)
where R is the circumradius.
Formula 5 — Inradius Formula:
Area = r × s
where r is the inradius and s is the semi-perimeter.
Special triangles:
- Equilateral triangle with side a: Area = (√3/4)a²
- Right triangle: Area = (1/2)(leg₁)(leg₂)
Useful relationships:
- Circumradius: R = a/(2 sin A) = b/(2 sin B) = c/(2 sin C)
- Inradius: r = Area / s
- For right triangle: hypotenuse = 2R (the hypotenuse is a diameter of the circumscribed circle)
Practical board exam strategy: If you have SAS data, use Formula 2 immediately. If you have SSS data, use Heron's formula. If you've already solved the triangle (all sides and angles known), use whichever formula is easiest to compute.
Applications
- Land surveying: computing the area of a triangulated lot or parcel.
- Structural design: computing tributary areas for load distribution.
- Highway design: computing earthwork cut/fill areas using cross-sections.
- Hydrology: computing watershed areas from surveyed triangulation points.
- Cost estimation: computing floor areas and material quantities for triangular roof panels.
Common misconceptions
- Using perimeter instead of semi-perimeter in Heron's formula — s = (a+b+c)/2, NOT (a+b+c).
- In Area = (1/2)ab sin C, using any angle instead of specifically the angle INCLUDED between sides a and b.
- Confusing the inradius r with the circumradius R — r is the inscribed circle radius (inside), R is the circumscribed circle radius (outside).
- For equilateral triangles, students sometimes try to compute height first unnecessarily — use Area = (√3/4)a² directly.
- Heron's formula applies only to triangles — it cannot be used for quadrilaterals or other polygons without subdivision.
Related concepts
- Law of Cosines
- Law of Sines
- Perimeter and Semi-perimeter
- Circumscribed and Inscribed Circles
- Coordinate Geometry Area Formula
Common exam questions
Example
Sides 3 m, 4 m, 5 m. s = 6. Area = √[6×3×2×1] = √36 = 6 m². (Note: 3-4-5 right triangle, check: (1/2)(3)(4) = 6 m² ✓)
Approach
Apply Heron's formula. Compute s first, then √[s(s−a)(s−b)(s−c)].
Example
a = 20 m, b = 25 m, C = 70°. Area = (1/2)(20)(25) sin70° = 250 × 0.9397 = 234.9 m².
Approach
Area = (1/2)ab sin C. No need to find the third side first.
Example
Equilateral triangle, side 10 m. Area = (√3/4)(100) = 43.30 m². s = 15. r = 43.30/15 = 2.89 m. R = 10/(2 sin60°) = 10/1.732 = 5.77 m. Note: R = 2r for equilateral triangle.
Approach
R = a/(2sinA) from Law of Sines. r = Area/s after computing Area and s.
Key points to remember
- Formula 2 (Area = (1/2)ab sin C) requires the INCLUDED angle — the angle BETWEEN the two given sides.
- Heron's formula s is the SEMI-perimeter: s = (a+b+c)/2, not (a+b+c).
- If the three sides give s(s−a)(s−b)(s−c) < 0, the triangle is impossible (violates triangle inequality).
- The circumradius R is found directly from the Law of Sines: R = a/(2 sin A).
- The inradius r = Area/s relates the area, inradius, and semi-perimeter — useful in competition-style problems.
- For equilateral triangle: Area = (√3/4)a² — memorize this for quick computation.
- Area computed by different formulas must give the same result — use this as a verification check.
Spherical Trigonometry
Examples
A spherical triangle has angles A = 95°, B = 85°, C = 100° on a sphere of radius 6 m. Find the spherical excess and the area of the triangle.
Solution
Step 1 — Compute spherical excess:
E = A + B + C − 180°
E = 95° + 85° + 100° − 180°
E = 280° − 180° = 100°
Step 2 — Compute area:
Area = (πR²E) / 180°
Area = (π × 6² × 100°) / 180°
Area = (π × 36 × 100) / 180
Area = 11,309.73 / 180
Area = 62.83 m²
Alternative: In radians, E = 100° × π/180 = 1.7453 rad
Area = E × R² = 1.7453 × 36 = 62.83 m² ✓
This is the standard board exam format for spherical trigonometry. The spherical excess formula E = (A+B+C) − 180° and Area = πR²E/180° must be memorized. Note that E = 100° means the triangle is quite large relative to the sphere.
In a spherical triangle, a = 70°, b = 50°, C = 80°. Find side c.
Solution
Step 1 — Apply the spherical law of cosines for sides:
cos c = cos a cos b + sin a sin b cos C
Step 2 — Substitute values:
cos c = cos 70° cos 50° + sin 70° sin 50° cos 80°
cos c = (0.3420)(0.6428) + (0.9397)(0.7660)(0.1736)
cos c = 0.2198 + 0.1250
cos c = 0.3448
Step 3 — Find c:
c = arccos(0.3448) = 69.83°
The spherical law of cosines for sides is the key formula for SAS-type problems in spherical trigonometry. Note the PLUS sign connecting the two terms — this distinguishes it from the plane law of cosines.
A spherical triangle is formed on the surface of a sphere by three great circle arcs. The sides of a spherical triangle are measured as angles (in degrees or radians), specifically as the central angles subtended at the center of the sphere by each arc.
Notation:
- Sides: a, b, c (in angular measure, degrees)
- Angles: A, B, C (interior angles at each vertex)
- The sphere has radius R
Key difference from plane triangles:
- In a plane triangle: A + B + C = 180° always.
- In a spherical triangle: A + B + C > 180° always. The excess is called SPHERICAL EXCESS E.
Spherical Excess:
E = (A + B + C) − 180°
Area of a spherical triangle:
Area = (πR² × E) / 180° [when E is in degrees]
or equivalently: Area = E × R² [when E is in radians]
Law of Sines for Spherical Triangles:
sin a / sin A = sin b / sin B = sin c / sin C
Law of Cosines for Sides (Spherical):
cos a = cos b cos c + sin b sin c cos A
cos b = cos a cos c + sin a sin c cos B
cos c = cos a cos b + sin a sin b cos C
Law of Cosines for Angles (Spherical) — Polar/Supplemental:
cos A = −cos B cos C + sin B sin C cos a
Right Spherical Triangle (Napier's Rules):
For a right spherical triangle with right angle at C:
The five circular parts (in order, skipping the right angle C) are: a, b, co-A, co-c, co-B
(where 'co-X' means complement of X = 90° − X)
Napier's rules:
- Sin of middle part = product of cosines of opposite parts
- Sin of middle part = product of tangents of adjacent parts
Applications in engineering:
- Geodesy: computing great-circle distances and bearings between geographic points
- Astronomy: celestial coordinate transformations
- Navigation: great-circle routes for ships and aircraft
- GPS and satellite positioning systems
Applications
- Geodesy and surveying: computing great-circle distances between cities for route planning.
- Navigation: determining true azimuth and great-circle bearing for ship navigation.
- Astronomy: converting between equatorial and horizontal coordinate systems for celestial observations.
- GPS and GNSS: satellite geometry and positioning calculations.
- Satellite communications: computing antenna pointing angles for ground stations.
Common misconceptions
- Thinking that A + B + C = 180° for spherical triangles — it is ALWAYS greater than 180°, never equal.
- Using the PLANE law of cosines (c² = a² + b² − 2ab cos C) for spherical triangles — the spherical version uses cos c = cos a cos b + sin a sin b cos C (angles in degrees, not side lengths).
- Assuming sides of a spherical triangle are lengths in meters — they are ANGULAR measures in degrees. Convert to arc length as: arc length = R × (angle in radians).
- Confusing spherical excess E with the sum A+B+C — E is the EXCESS above 180°.
- Napier's Rules only apply to RIGHT spherical triangles — do not use them for oblique spherical triangles.
Related concepts
- Great Circles and Geodesics
- Plane Trigonometry (Law of Sines and Cosines)
- Geodesy and Surveying
- Geographic Coordinate System
- Solid Geometry (Spheres)
Common exam questions
Example
Spherical triangle on R = 10 m sphere: A = 110°, B = 90°, C = 85°. E = 285° − 180° = 105°. Area = π(100)(105)/180 = 183.26 m².
Approach
E = A + B + C − 180°. Area = πR²E/180° (E in degrees). Straightforward substitution.
Example
b = 60°, c = 45°, A = 90°. cos a = cos60°cos45° + sin60°sin45°cos90° = 0.5×0.7071 + (0.8660×0.7071×0) = 0.3536. a = arccos(0.3536) = 69.30°.
Approach
cos a = cos b cos c + sin b sin c cos A. Identify known quantities and solve for the unknown.
Example
Right spherical triangle C = 90°, a = 40°, b = 50°. Circular parts: a, b, co-A, co-c, co-B. To find co-c (i.e., find c): sin(co-c) = cos(a)×cos(b) → cos c = cos40°cos50° = 0.7660×0.6428 = 0.4924. c = arccos(0.4924) = 60.48°.
Approach
Arrange circular parts in order (omitting the right angle). Apply: sin(middle) = cos(opposite)×cos(opposite) or sin(middle) = tan(adjacent)×tan(adjacent).
Key points to remember
- In a spherical triangle, all sides AND all angles are measured in degrees (angular measure). Sides are not lengths but central angles.
- A + B + C > 180° always for spherical triangles — this is the fundamental difference from plane trigonometry.
- Spherical excess E = (A + B + C) − 180° and Area = πR²E/180° (E in degrees).
- The spherical law of cosines uses cos a = cos b cos c + sin b sin c cos A — notice the PLUS sign before sin b sin c cos A (different from plane version).
- For small spherical triangles (small sides), spherical trig approaches plane trig — the difference becomes negligible for small areas.
- Napier's circular rules apply specifically to RIGHT spherical triangles — very useful for reducing computation.
- Board exams typically test spherical excess and area calculation for given angle values.
Practice problems
Problem 1 (AAS — Law of Sines): In triangle ABC, angle A = 35°, angle B = 65°, and side a = 18 m (opposite to A). Find sides b and c, and the area of the triangle.
Solution
Step 1 — Find angle C:
C = 180° − 35° − 65° = 80°
Step 2 — Find side b using Law of Sines:
b / sin B = a / sin A
b = a × sin B / sin A
b = 18 × sin 65° / sin 35°
b = 18 × 0.9063 / 0.5736
b = 28.42 m
Step 3 — Find side c:
c = a × sin C / sin A
c = 18 × sin 80° / sin 35°
c = 18 × 0.9848 / 0.5736
c = 30.89 m
Step 4 — Area:
Area = (1/2) × b × c × sin A
Area = (1/2)(28.42)(30.89) × sin 35°
Area = (1/2)(28.42)(30.89)(0.5736)
Area = 252.13 m²
Verify: Area = (1/2)(a)(b) sin C = (1/2)(18)(28.42) sin 80° = 251.9 ≈ 252 m² ✓
Classic AAS setup. Find the missing angle first, then use Law of Sines ratios. Use any two adjacent sides with their included angle for area computation. Small rounding differences in the verification are acceptable in board exam settings.
Problem 2 (SSA — Ambiguous Case): Given a = 9 m, b = 12 m, A = 28°. Determine the number of valid triangles and solve completely for each.
Solution
Step 1 — Check the case: A = 28° < 90°, so check height:
h = b sin A = 12 × sin 28° = 12 × 0.4695 = 5.63 m
Step 2 — Compare a to h and b:
h = 5.63 m, a = 9 m, b = 12 m
Since h (5.63) < a (9) < b (12): TWO triangles exist.
Step 3 — Find B using Law of Sines:
sin B = b sin A / a = 12 × 0.4695 / 9 = 0.6260
B₁ = arcsin(0.6260) = 38.77°
B₂ = 180° − 38.77° = 141.23°
Step 4 — Verify B₂: A + B₂ = 28° + 141.23° = 169.23° < 180° ✓ (B₂ is valid)
Triangle 1 (acute B):
C₁ = 180° − 28° − 38.77° = 113.23°
c₁ = a sin C₁ / sin A = 9 × sin(113.23°) / sin(28°)
c₁ = 9 × 0.9191 / 0.4695 = 17.62 m
Area₁ = (1/2)(9)(17.62) sin 28° = (1/2)(9)(17.62)(0.4695) = 37.27 m²
Triangle 2 (obtuse B):
C₂ = 180° − 28° − 141.23° = 10.77°
c₂ = a sin C₂ / sin A = 9 × sin(10.77°) / sin(28°)
c₂ = 9 × 0.1868 / 0.4695 = 3.58 m
Area₂ = (1/2)(9)(3.58) sin 28° = (1/2)(9)(3.58)(0.4695) = 7.57 m²
The ambiguous case requires systematic checking of h vs. a vs. b. Board exams may ask 'how many triangles' as a standalone question, or may require solving both. Always present both triangles clearly labeled.
Problem 3 (SAS — Law of Cosines): A surveyor measures two sides of a triangular lot as 120 m and 180 m with an included angle of 110°. Find the third side and the area of the lot.
Solution
Step 1 — Apply Law of Cosines:
Let a = 120 m, b = 180 m, C = 110°
c² = a² + b² − 2ab cos C
c² = 120² + 180² − 2(120)(180) cos 110°
c² = 14,400 + 32,400 − 43,200 × (−0.3420)
c² = 46,800 + 14,774.4
c² = 61,574.4
c = √61,574.4 = 248.14 m
Step 2 — Area:
Area = (1/2) × a × b × sin C
Area = (1/2)(120)(180) sin 110°
Area = 10,800 × 0.9397
Area = 10,148.8 m²
Area ≈ 10,149 m² ≈ 1.0149 hectares
Note that cos 110° is NEGATIVE (obtuse angle), so the −2ab cos C term becomes POSITIVE (subtracting a negative). This causes c to be larger than either a or b, which makes physical sense for an obtuse angle. The area formula (1/2)ab sin C works directly — sin 110° is positive.
Problem 4 (SSS — All three sides): A triangle has sides of 13 m, 14 m, and 15 m. Find all three angles and the area using Heron's formula.
Solution
Step 1 — Heron's formula:
s = (13 + 14 + 15) / 2 = 21 m
Area = √[21(21−13)(21−14)(21−15)]
= √[21 × 8 × 7 × 6]
= √[7,056]
= 84 m²
Step 2 — Find the largest angle first (opposite side 15 m, call it C):
cos C = (a² + b² − c²) / (2ab)
cos C = (13² + 14² − 15²) / (2 × 13 × 14)
cos C = (169 + 196 − 225) / 364
cos C = 140 / 364 = 0.3846
C = arccos(0.3846) = 67.38°
Step 3 — Find angle A (opposite side 13 m):
cos A = (14² + 15² − 13²) / (2 × 14 × 15)
cos A = (196 + 225 − 169) / 420
cos A = 252 / 420 = 0.6000
A = arccos(0.6000) = 53.13°
Step 4 — Find angle B by subtraction:
B = 180° − 67.38° − 53.13° = 59.49°
Verify: (1/2)(13)(14) sin 67.38° = 91 × 0.9231 = 84 m² ✓
The 13-14-15 triangle is a classic exam problem with clean numbers. Heron's formula gives an integer area (84 m²), which is a hint that the numbers work out nicely. Verification via the formula (1/2)ab sin C confirms the computation.
Problem 5 (Angle of Elevation — Applied): From a point P on level ground, the angle of elevation to the top of a 60-m vertical tower is 42°. A second point Q is on the other side of the tower (collinear with P and the base of the tower) at a distance of 80 m from P. Find the angle of elevation to the top of the tower from Q.
Solution
Step 1 — Find horizontal distance from P to the tower base:
Let d = distance from P to the tower base.
tan 42° = 60 / d
d = 60 / tan 42° = 60 / 0.9004 = 66.64 m
Step 2 — Find distance from Q to the tower base:
Since P, base, and Q are collinear, and P and Q are on opposite sides:
Distance from Q to base = 80 − 66.64 = 13.36 m
(Since total P-to-Q = 80 m and P-to-base = 66.64 m, base is between P and Q)
Step 3 — Find angle of elevation from Q:
tan θ = 60 / 13.36 = 4.491
θ = arctan(4.491) = 77.46°
Answer: The angle of elevation from Q is 77.46°.
Angle of elevation problems require careful drawing of the geometry. Here, the tower base lies between P and Q, so the distance from Q to the tower base is 80 − 66.64 = 13.36 m. This type of problem tests both the basic trig function application and the geometric setup.
Problem 6 (Spherical Trigonometry): Two points on Earth have the following coordinates: Point A at latitude 14.6°N, longitude 121.0°E (Manila) and Point B at latitude 35.7°N, longitude 139.7°E (Tokyo). Compute the great-circle angular distance between them using the spherical law of cosines. Take Earth's radius as 6,371 km and find the actual great-circle distance.
Solution
Step 1 — Set up the spherical triangle.
In spherical navigation, use the formula:
cos c = sin(lat_A) sin(lat_B) + cos(lat_A) cos(lat_B) cos(Δlong)
where Δlong = difference in longitude.
Step 2 — Compute Δlong:
Δlong = 139.7° − 121.0° = 18.7°
Step 3 — Apply the formula:
cos c = sin(14.6°) sin(35.7°) + cos(14.6°) cos(35.7°) cos(18.7°)
= (0.2521)(0.5835) + (0.9677)(0.8121)(0.9469)
= 0.1470 + 0.7443
= 0.8913
Step 4 — Find the angular distance c:
c = arccos(0.8913) = 26.90°
Step 5 — Convert to actual distance:
Distance = R × c (in radians)
Distance = 6,371 × (26.90° × π / 180°)
Distance = 6,371 × 0.4695
Distance = 2,991 km
This is the standard great-circle distance formula derived from the spherical law of cosines. In navigation applications, latitudes are co-latitudes measured from the North Pole: a = 90° − lat_A, b = 90° − lat_B, and C = Δlong. The formula simplifies to the form shown above. The actual Manila-Tokyo great-circle distance is approximately 2,990 km, consistent with our result.
Exam preparation tips
- CALCULATOR DISCIPLINE: Set your scientific calculator to DEGREE mode at the start of every problem. Re-check after every reset. A radians/degrees error will give a completely wrong answer with no way to detect it from the problem setup.
- SOH-CAH-TOA vs. LAW OF SINES/COSINES: Right triangle → SOH-CAH-TOA always. Oblique triangle (no right angle) → choose Law of Sines (AAS, ASA, SSA) or Law of Cosines (SAS, SSS). Never mix the two approaches.
- SSA AMBIGUOUS CASE DRILL: Memorize the decision tree — (1) Is A ≥ 90°? One or no triangle. (2) If A < 90°, compute h = b sin A. Compare a vs. h: a < h → no triangle, a = h → one right triangle, h < a < b → TWO triangles, a ≥ b → one triangle. Expect 1–2 board questions on this every examination.
- LAW OF COSINES ANGLE: Always double-check that the angle in cos C is the angle BETWEEN sides a and b. A mislabeled angle gives a completely wrong side length. Write out the formula and label each variable before substituting.
- HERON'S FORMULA — SEMI-PERIMETER PITFALL: s = (a+b+c)/2. The most common error is using the full perimeter. Write 's = half of (a+b+c)' explicitly in your solution.
- LARGEST ANGLE FIRST IN SSS: When solving for angles from three sides, find the largest angle (opposite the longest side) using Law of Cosines. This guarantees you correctly identify an obtuse angle. The remaining angles found by subtraction or Law of Sines will necessarily be acute.
- DOUBLE-ANGLE FORMULA CHOICE: For cos 2θ, choose the form that eliminates the unknown: if only sin θ is given, use cos 2θ = 1 − 2sin²θ; if only cos θ is given, use cos 2θ = 2cos²θ − 1; if both are known, use cos 2θ = cos²θ − sin²θ.
- SPHERICAL TRIGONOMETRY MUST-KNOWS: E = (A+B+C) − 180° and Area = πR²E/180°. These two formulas cover the majority of spherical trig board questions. Know them cold.
- ANGLE SUM VERIFICATION: After solving any triangle, ALWAYS add all three angles. If A + B + C ≠ 180° (within rounding tolerance of ±0.5°), you made an error somewhere. This takes 5 seconds and can save you from marking a wrong answer.
- AREA CONSISTENCY CHECK: Compute area using two different formulas when possible. If both give the same result (within rounding), your side and angle calculations are consistent and likely correct.
- PRACTICE WITH BOARD-STYLE NUMBERS: Philippine board exams typically use 'clean' angles (30°, 45°, 60°, specific degree values) or specific integer/simple fraction side lengths. Practice with these types to develop speed.
- TIME MANAGEMENT: Trigonometry problems in the board exam average 2–4 minutes each. If a problem is taking too long, move on and return. Complex ambiguous-case or chain problems should be attempted last within the mathematics section.
- FORMULA SHEET AWARENESS: You cannot bring formula sheets to the board exam. Prioritize memorizing: (1) Pythagorean identities, (2) sum/difference and double-angle formulas, (3) Law of Sines, (4) Law of Cosines, (5) Heron's formula, (6) Area = (1/2)ab sin C, (7) Spherical excess and area.
Diagrams
Plane and Spherical Trigonometry — Complete Concept Map
A comprehensive mind map showing all major topics in Plane and Spherical Trigonometry, branching from the core concept outward to right triangle functions, identities, oblique triangle laws, and spherical trigonometry. Use this for quick review and to see how all topics interconnect.
Decision Tree: Choosing the Right Formula for Oblique Triangle Problems
An exam-strategy flowchart that guides you through selecting the correct formula based on the given data (AAS, ASA, SAS, SSS, or SSA). Follow the decision tree from top to bottom: identify what is given, check the case, apply the appropriate law, and compute the area. The ambiguous SSA case is highlighted with warning icons.
Step-by-Step Solution Procedure: Law of Cosines (SAS and SSS)
A detailed procedural flowchart for solving triangles using the Law of Cosines. Two parallel tracks show the SAS procedure (left) and SSS procedure (right). A critical decision node highlights when cosine is negative (obtuse angle). Always ends with angle-sum verification and area computation.
Trigonometric Identities — Key Relationships and Derivation Path
A flowchart showing how ALL major trigonometric identities derive from the single master identity sin²θ + cos²θ = 1. Follow the arrows to see how Pythagorean identities are derived by dividing, how double-angle formulas follow from sum formulas with A=B, and how half-angle formulas come from the double-angle forms.
Distribution of Trigonometry Question Types in PRC Civil Engineer Board Exams
A pie chart showing the approximate distribution of trigonometry-related question types that appear in the PRC Civil Engineer Licensure Examination. Law of Sines and Law of Cosines together account for approximately 42% of trig questions. Prioritize these two laws along with identities and angle of elevation/depression problems for maximum exam coverage.
Spherical Trigonometry — Solution Procedure and Key Formulas
A solution-guide flowchart for spherical trigonometry problems. Starting from the problem type (given angles only, SAS, SSA, or right spherical triangle), it directs you to the appropriate formula. Critical reminders are shown at the bottom: sides are angular measures in degrees, and the sum of angles always exceeds 180°.
In summary
Plane and Spherical Trigonometry is a high-yield chapter in the PRC Civil Engineer Licensure Examination that rewards systematic preparation. The subject is highly structured: a small set of core formulas (Law of Sines, Law of Cosines, and the major identities) governs the vast majority of exam problems. Your examination success in this chapter hinges on three competencies: (1) formula mastery — knowing exactly which formula to apply for each given set of information, using the decision-tree logic practiced in this chapter; (2) computational accuracy — particularly careful substitution into the Law of Cosines where sign errors are common, and correct use of degree mode on your calculator; and (3) awareness of pitfalls — especially the SSA ambiguous case, the semi-perimeter in Heron's formula, the largest-angle-first strategy in SSS problems, and the fundamental difference between plane and spherical triangle angle sums.
For spherical trigonometry, the two formulas E = (A+B+C) − 180° and Area = πR²E/180° cover most board questions. For plane trigonometry, the four most productive study activities are: (1) drilling the special angle values (30°, 45°, 60°) until they are automatic; (2) solving complete triangle problems from all five cases (AAS, ASA, SSA, SAS, SSS) until the decision process is second nature; (3) practicing identity simplification using the Pythagorean and double-angle identities; and (4) solving applied problems (angles of elevation and depression, bearing problems, and area computations) that integrate multiple steps.
Remember: in the board examination hall, these problems are time-constrained. Practice not just for correctness, but for speed. With consistent daily drill using worked examples at the licensure-review level, you will confidently solve trigonometry problems within the allotted time and maximize your score in Engineering Mathematics.