GELE Adjustment Computations (Least Squares) — Condition Equations and Figure AdjustmentExam Answer Templates

Marks

1

Topic

Condition Equations — Definition

Difficulty

easy

Template Id

T1

Examiner Tip

One crisp sentence with 'adjusted observations', 'exactly', and one example earns full mark. Do not over-explain.

Model Answer

A condition equation is a mathematical relationship that the adjusted observations in a geodetic figure must satisfy exactly, expressing a known geometric constraint such as the sum of angles in a plane triangle equalling 180°.

Question Type

very_short_answer

Answer Structure

• One sentence: define condition equation with the key phrase 'must satisfy exactly' and cite one geometric example [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing 'observed observations' instead of 'adjusted observations' — the condition applies to the adjusted, not raw, values.
• Omitting any example of a geometric condition — examiners expect at least one instance.
• Confusing condition equation with observation equation.

Key Phrases To Include

• adjusted observations
• must satisfy exactly
• geometric constraint
• known relationship

Marks

1

Topic

Condition Equations — Misclosure

Difficulty

easy

Template Id

T2

Examiner Tip

The formula is the mark-earner here. Write it explicitly even if you explain it in words.

Model Answer

Misclosure is the amount by which the sum of the raw (unadjusted) observations violates the required geometric condition. It is computed as: misclosure = Σ(observed) − required total.

Question Type

very_short_answer

Answer Structure

• One sentence: define misclosure as the violation of the geometric condition [0.5 mark]
• One formula: misclosure = Σ(observed) − required total [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Writing the formula with the wrong sign (required total − observed) without defining the sign convention.
• Omitting the formula — definition alone scores only partial credit.

Key Phrases To Include

• misclosure
• violation
• Σ(observed) − required total
• unadjusted observations

Marks

1

Topic

Equal Distribution — Correction Formula

Difficulty

easy

Template Id

T3

Examiner Tip

The negative sign is everything in this formula. Box it or underline it in your answer.

Model Answer

For n equally-weighted observations, the correction applied to each observation is: c = −misclosure / n. The correction is equal in magnitude for all observations and opposite in sign to the misclosure.

Question Type

very_short_answer

Answer Structure

• Formula: c = −misclosure / n [0.5 mark]
• Statement: correction is opposite in sign to the misclosure [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Omitting the negative sign — this is the most common error and always results in mark deduction.
• Writing c = misclosure / n without the negative sign.

Key Phrases To Include

• c = −misclosure / n
• equal weights
• opposite in sign
• distributed equally

Marks

2

Topic

Plane vs. Spherical Triangle Condition

Difficulty

medium

Template Id

T4

Examiner Tip

The phrase '180° + ε' written clearly earns you the key mark. Examiners specifically look for this in any spherical triangle question.

Model Answer

Plane triangle condition: The sum of the three interior angles of a plane triangle must equal exactly 180°00'00''. Required total = 180°. Spherical triangle condition: Because the triangle is on a curved surface, the sum of the three interior angles must equal 180° plus the spherical excess (ε), where ε = area of triangle / R². Required total = 180° + ε. The spherical excess ε is significant for large triangles (sides > 10 km) as encountered in first-order geodetic triangulation in the Philippines under PRS92.

Question Type

short_answer

Answer Structure

• Point 1: State plane triangle required total = 180° [0.5 mark]
• Point 2: State spherical triangle required total = 180° + ε [0.5 mark]
• Point 3: Define spherical excess ε [0.5 mark]
• Point 4: Note when spherical excess is significant (large triangles / geodetic work) [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using 180° as the required total for a spherical triangle — this loses 1 mark immediately.
• Failing to define ε or explain when it matters.
• Confusing spherical excess with angular misclosure.

Key Phrases To Include

• required total = 180°
• spherical excess ε
• 180° + ε
• curved surface
• geodetic triangulation

Marks

2

Topic

Horizon Condition Adjustment

Difficulty

medium

Template Id

T5

Examiner Tip

Always write the CHECK line. Even if your adjusted values are wrong, a correct verification attempt shows methodology and can earn partial credit.

Model Answer

Given: A₁ = 119°59'54'', A₂ = 120°00'10'', A₃ = 120°00'02'' Condition: Horizon condition — angles around a point must sum to 360°00'00'' Required total = 360°00'00'' Step 1 — Compute misclosure: Σ(observed) = 119°59'54'' + 120°00'10'' + 120°00'02'' = 360°00'06'' Misclosure = 360°00'06'' − 360°00'00'' = +6'' Step 2 — Compute correction (equal weights, n = 3): c = −misclosure / n = −6'' / 3 = −2'' per angle Step 3 — Apply corrections: A₁ (adjusted) = 119°59'54'' + (−2'') = 119°59'52'' A₂ (adjusted) = 120°00'10'' + (−2'') = 120°00'08'' A₃ (adjusted) = 120°00'02'' + (−2'') = 120°00'00'' CHECK: 119°59'52'' + 120°00'08'' + 120°00'00'' = 360°00'00'' ✓

Question Type

numerical

Answer Structure

• Line 1: Identify condition type (horizon, required total = 360°) [0 mark — setup]
• Line 2: Compute Σ(observed) and misclosure [0.5 mark]
• Line 3: Apply correction formula c = −misclosure/n [0.5 mark]
• Line 4: State all three adjusted angles [0.5 mark]
• Line 5: Verification check [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Adding +2'' instead of −2'' (wrong sign of correction).
• Omitting the verification check — loses 0.5 mark.
• Arithmetic error in summing DMS angles — convert carefully.

Key Phrases To Include

• horizon condition
• required total = 360°00'00''
• misclosure = +6''
• c = −2'' per angle
• CHECK: Σadjusted = 360°00'00'' ✓

Marks

3

Topic

Triangle Angle Condition — Equal Weights

Difficulty

medium

Template Id

T6

Examiner Tip

DMS arithmetic is the most common error source. Write it out column by column: degrees, minutes, seconds — then carry over. Never rush this step.

Model Answer

Given: A = 58°20'30'', B = 61°40'18'', C = 60°00'06'' Condition: Plane triangle — Σangles = 180°00'00'' Step 1 — Compute Σ(observed): A + B + C = 58°20'30'' + 61°40'18'' + 60°00'06'' = (58+61+60)° + (20+40+00)' + (30+18+06)'' = 179° + 60' + 54'' = 179°60'54'' = 180°00'54'' Step 2 — Compute misclosure: Misclosure = 180°00'54'' − 180°00'00'' = +54'' Step 3 — Correction formula (equal weights, n = 3): c = −misclosure / n = −54'' / 3 = −18'' per angle Step 4 — Apply corrections: A (adjusted) = 58°20'30'' − 18'' = 58°20'12'' B (adjusted) = 61°40'18'' − 18'' = 61°40'00'' C (adjusted) = 60°00'06'' − 18'' = 59°59'48'' Step 5 — Verification: 58°20'12'' + 61°40'00'' + 59°59'48'' = 180°00'00'' ✓

Question Type

numerical

Answer Structure

• Step 1: Sum the observed angles correctly in DMS [0.5 mark]
• Step 2: Compute misclosure = +54'' [0.5 mark]
• Step 3: State and apply correction formula c = −18'' [0.5 mark]
• Step 4: State all three adjusted angles correctly [1 mark]
• Step 5: Verification check confirming 180°00'00'' [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Incorrect DMS addition — not carrying minutes to degrees properly (60' = 1°).
• Applying correction with wrong sign (adding instead of subtracting).
• Skipping the verification step.
• Not stating the condition equation at the beginning.

Key Phrases To Include

• plane triangle condition
• Σangles = 180°00'00''
• misclosure = +54''
• c = −18'' per angle
• verification ✓

Marks

3

Topic

Weighted Distribution

Difficulty

medium

Template Id

T7

Examiner Tip

The phrase 'weaker observations receive more correction' is a guaranteed mark-earner. Write it explicitly and early in your answer.

Model Answer

Principle of Weighted Distribution: When observations have unequal reliability, the misclosure is distributed in proportion to the relative variance (or inversely proportional to weight) of each observation — weaker observations (higher variance, lower weight) receive a larger share of the correction. Formulation: Let the total misclosure = W. For n observations with weights w₁, w₂, ..., wₙ: The correction to observation i is: cᵢ = −W × (1/wᵢ) / Σ(1/wⱼ) Rationale: A high-variance (low-weight) observation is less reliable; the adjustment assigns more correction to it because its true value is more uncertain, while a precise (high-weight) observation is disturbed as little as possible. Example context: In a Philippine first-order triangulation network under PRS92, angles measured with a 1'' theodolite have higher weight than those measured with a 6'' theodolite, and therefore receive a smaller correction when adjusting a triangle misclosure.

Question Type

short_answer

Answer Structure

• Sentence 1: State the principle — weaker observations (lower weight) receive more correction [1 mark]
• Equation: Write cᵢ = −W × (1/wᵢ) / Σ(1/wⱼ) [1 mark]
• Explanation: Justify why weaker observations get more correction [0.5 mark]
• Example: Apply to a geodetic context [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Stating that higher-weight observations receive more correction — this is the OPPOSITE of the correct principle.
• Omitting the formula and giving only a qualitative description.
• Confusing weight with variance (weight = 1/variance, not variance itself).

Key Phrases To Include

• proportional to variance
• inversely proportional to weight
• weaker observations receive more correction
• cᵢ = −W × (1/wᵢ) / Σ(1/wⱼ)
• high-weight observations disturbed least

Marks

3

Topic

Spherical Triangle Condition

Difficulty

hard

Template Id

T8

Examiner Tip

This problem is a deliberate trap: the angles already satisfy the spherical condition. Examiners reward students who recognise zero misclosure and state 'no correction needed' rather than inventing corrections.

Model Answer

Given: P = 72°15'22'', Q = 58°30'16'', R = 49°14'30'' Spherical excess: ε = 8'' Condition: Spherical triangle — required total = 180° + ε = 180°00'08'' Step 1 — Compute Σ(observed): P + Q + R = 72°15'22'' + 58°30'16'' + 49°14'30'' = (72+58+49)° + (15+30+14)' + (22+16+30)'' = 179° + 59' + 68'' = 179° + 60' + 8'' = 180°00'08'' Step 2 — Compute misclosure: Misclosure = 180°00'08'' − (180°00'08'') = 0'' Interpretation: The measured angles already satisfy the spherical triangle condition exactly. No correction is needed. Adjusted angles: P = 72°15'22'', Q = 58°30'16'', R = 49°14'30'' (unchanged) CHECK: 72°15'22'' + 58°30'16'' + 49°14'30'' = 180°00'08'' = 180° + 8'' ✓

Question Type

numerical

Answer Structure

• Step 1: State required total = 180° + ε = 180°00'08'' [1 mark]
• Step 2: Sum observed angles correctly = 180°00'08'' [0.5 mark]
• Step 3: Compute misclosure = 0'' [0.5 mark]
• Step 4: Conclude no correction needed and state final adjusted angles [0.5 mark]
• Step 5: Verification check [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using 180° instead of 180° + ε as the required total — this is the most common and most heavily penalised error in spherical triangle problems.
• Computing misclosure against 180° (getting −8'') and distributing a correction that should not exist.
• Arithmetic error in DMS addition.

Key Phrases To Include

• required total = 180° + ε
• 180°00'08''
• misclosure = 0''
• spherical excess
• first-order triangulation

Marks

3

Topic

Loop Closure Condition — Levelling

Difficulty

medium

Template Id

T9

Examiner Tip

Mentioning the NAMRIA permissible closure standard (12√K mm for third-order) adds professional credibility and often earns the 'bonus' mark from examiners.

Model Answer

Procedure for Adjusting a Closed Levelling Loop: Step 1 — Identify the condition: The loop closure condition requires that the algebraic sum of all observed elevation differences around the loop equals zero: Σ(ΔH_observed) = 0. Step 2 — Compute the misclosure: Misclosure = Σ(ΔH_observed) − 0 = Σ(ΔH_observed) A positive misclosure means the loop closed high; negative means closed low. Step 3 — Apply corrections: For equal-reliability sections: c = −misclosure / n (per section). For unequal reliability (different section lengths): corrections are proportional to the length of each section: cᵢ = −misclosure × (Lᵢ / ΣLⱼ) where Lᵢ = length of section i. Step 4 — Compute adjusted elevations: Adjusted ΔHᵢ = Observed ΔHᵢ + cᵢ Step 5 — Verify: Σ(Adjusted ΔH) = 0 ✓ Note: In the Philippines, third-order levelling loops under NAMRIA standards have a permissible closure of 12√K mm, where K is the loop length in kilometres.

Question Type

short_answer

Answer Structure

• Step 1: State the loop closure condition Σ(ΔH) = 0 [0.5 mark]
• Step 2: Define and compute misclosure [0.5 mark]
• Step 3: Give correction formula for equal and length-weighted cases [1 mark]
• Step 4: State how adjusted values are computed [0.5 mark]
• Step 5: Verification and Philippine standard reference [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using equal correction per section when sections have different lengths — equal weighting is only valid for equal-length sections.
• Omitting the verification step.
• Confusing loop misclosure (elevation closure) with a traverse angular misclosure.

Key Phrases To Include

• loop closure condition
• Σ(ΔH) = 0
• misclosure
• proportional to section length
• cᵢ = −misclosure × (Lᵢ / ΣLⱼ)
• verification

Marks

5

Topic

Weighted Distribution — Unequal Reliability

Difficulty

hard

Template Id

T10

Examiner Tip

In weighted problems, ALWAYS write 'weight is inversely proportional to variance' as your first line. Then assign numbers. If you get the weights wrong, everything downstream is wrong — so this line is worth fighting for.

Model Answer

Given: Misclosure W = +9'' Angles A and B: equal variance = σ² Angle C: variance = 2σ² (twice the variance — less reliable) Step 1 — Assign weights: Weight is inversely proportional to variance. wA = 1/σ² → let wA = 1 wB = 1/σ² → wB = 1 wC = 1/(2σ²) → wC = 0.5 Step 2 — Compute reciprocals of weights (proportional to variance): 1/wA = 1, 1/wB = 1, 1/wC = 2 Σ(1/wⱼ) = 1 + 1 + 2 = 4 Step 3 — Compute corrections: cA = −W × (1/wA) / Σ(1/wⱼ) = −9'' × (1/4) = −2.25'' cB = −W × (1/wB) / Σ(1/wⱼ) = −9'' × (1/4) = −2.25'' cC = −W × (1/wC) / Σ(1/wⱼ) = −9'' × (2/4) = −4.50'' Step 4 — Verify total correction = −W: cA + cB + cC = −2.25'' + (−2.25'') + (−4.50'') = −9.00'' = −W ✓ Step 5 — Summary of corrections: Angle A: correction = −2.25'' (≈ −2'') Angle B: correction = −2.25'' (≈ −2'') Angle C: correction = −4.50'' (≈ −5'') Conclusion: Angle C, being twice as uncertain, receives twice the correction of angles A or B. The total misclosure is fully absorbed. (Note: If integer seconds are required, round −2.25'' to −2'' for A and B, and −4.50'' to −5'' for C, giving −2−2−5 = −9'' ✓.)

Question Type

numerical

Answer Structure

• Step 1: Define weights from variances — wA = wB = 1, wC = 0.5 [1 mark]
• Step 2: Compute 1/wᵢ for each angle and their sum [1 mark]
• Step 3: Apply weighted correction formula for each angle [1 mark]
• Step 4: Verify corrections sum to −W [1 mark]
• Step 5: State final corrections clearly and interpret [1 mark]

Scoring Breakdown

Common Mark Deductions

• Assigning wC = 2 instead of wC = 0.5 — confusing weight with variance is the most common conceptual error.
• Giving angle C a smaller correction than A or B — the physically wrong answer.
• Omitting the verification step (loses 1 mark).
• Not justifying the weight assignments from the variance relationship.
• Using the equal-weight formula for this problem.

Key Phrases To Include

• weight inversely proportional to variance
• wC = 0.5
• 1/wC = 2
• cᵢ = −W × (1/wᵢ) / Σ(1/wⱼ)
• weaker observation receives more correction
• verification: ΣcI = −W ✓

Marks

5

Topic

Triangle Angle Condition — Quadrilateral

Difficulty

medium

Template Id

T11

Examiner Tip

Write DMS addition line by line — degrees column, minutes column, seconds column — with carry-overs shown explicitly. This approach earns the method mark even if you make one arithmetic slip.

Model Answer

Given: Triangle ABD (plane triangle) Angle A = 45°00'08'' Angle B = 90°00'04'' Angle D = 44°59'54'' Condition: Σangles = 180°00'00'' Step 1 — Compute Σ(observed): Σ = 45°00'08'' + 90°00'04'' + 44°59'54'' Degrees: 45 + 90 + 44 = 179° Minutes: 00 + 00 + 59 = 59' Seconds: 08 + 04 + 54 = 66'' = 1'06'' Σ = 179° + 59' + 66'' = 179° + 60' + 6'' = 180°00'06'' Step 2 — Compute misclosure: Misclosure = 180°00'06'' − 180°00'00'' = +6'' Step 3 — Correction (equal weights, n = 3): c = −6'' / 3 = −2'' per angle Step 4 — Apply corrections: A (adjusted) = 45°00'08'' − 2'' = 45°00'06'' B (adjusted) = 90°00'04'' − 2'' = 90°00'02'' D (adjusted) = 44°59'54'' − 2'' = 44°59'52'' Step 5 — Verification: 45°00'06'' + 90°00'02'' + 44°59'52'' = (45+90+44)° + (00+00+59)' + (06+02+52)'' = 179° + 59' + 60'' = 179° + 60' = 180°00'00'' ✓ Conclusion: The three adjusted angles of triangle ABD sum to exactly 180°00'00'', satisfying the plane triangle condition.

Question Type

numerical

Answer Structure

• Step 1: Correct DMS summation of three angles [1 mark]
• Step 2: Correct misclosure = +6'' [1 mark]
• Step 3: Correct correction formula and value c = −2'' [1 mark]
• Step 4: All three adjusted angles stated correctly [1 mark]
• Step 5: Full verification with DMS arithmetic [1 mark]

Scoring Breakdown

Common Mark Deductions

• DMS carry-over error: writing 66'' without converting to 1'06'' causes a 1° error in the sum.
• Not showing the DMS addition step — examiners need to see the arithmetic.
• Forgetting to verify the final adjusted values.
• Rounding 2'' to 0 — all corrections must be applied, even small ones.

Key Phrases To Include

• plane triangle condition
• Σ = 180°00'06''
• misclosure = +6''
• c = −2'' per angle
• adjusted sum = 180°00'00'' ✓

Marks

3

Topic

Types of Condition Equations

Difficulty

easy

Template Id

T12

Examiner Tip

Each condition must have its mathematical formula — a description without the formula earns only half the mark per condition.

Model Answer

The three most common condition equations in geodetic figure adjustment are: 1. Triangle (Angle) Condition: The sum of the three interior angles of a plane triangle must equal 180°00'00''. For spherical triangles (large geodetic triangles), the required total is 180° + ε, where ε is the spherical excess. Formula: A + B + C = 180° (plane) or 180° + ε (spherical) 2. Horizon (Station) Condition: All angles observed around a survey station must sum to 360°00'00'', since they constitute a full revolution around the point. Formula: Σ(angles around station) = 360° 3. Loop (Closure) Condition: In a closed traverse or levelling loop, the sum of the observed quantity (elevation differences or coordinate increments) must return to its starting value — i.e., the algebraic sum equals zero. Formula: Σ(ΔH) = 0 (levelling loop) or Σ(ΔE) = Σ(ΔN) = 0 (closed traverse)

Question Type

short_answer

Answer Structure

• Condition 1: Triangle condition — correct required total for plane and spherical [1 mark]
• Condition 2: Horizon condition — correct required total 360° [1 mark]
• Condition 3: Loop closure condition — correct formula [1 mark]

Scoring Breakdown

Common Mark Deductions

• Describing only one or two types instead of all three.
• Omitting the spherical case for the triangle condition.
• Stating the horizon condition requires 180° instead of 360°.
• Not providing the mathematical formula for any condition.

Key Phrases To Include

• triangle condition
• 180° + ε
• horizon condition
• 360°
• loop closure condition
• Σ(ΔH) = 0

Marks

5

Topic

Horizon Condition — Four Angles

Difficulty

medium

Template Id

T13

Examiner Tip

For four-angle horizon problems, many students use n = 3 by mistake. Count the angles explicitly in your solution and write 'n = 4' clearly.

Model Answer

Given: α₁ = 88°00'05'', α₂ = 92°00'03'', α₃ = 88°59'58'', α₄ = 91°00'02'' Condition: Horizon condition — Σ = 360°00'00'' Step 1 — Compute Σ(observed): Degrees: 88 + 92 + 88 + 91 = 359° Minutes: 00 + 00 + 59 + 00 = 59' Seconds: 05 + 03 + 58 + 02 = 68'' = 1'08'' Σ = 359° + 59' + 68'' = 359° + 60' + 8'' = 360°00'08'' Step 2 — Compute misclosure: Misclosure = 360°00'08'' − 360°00'00'' = +8'' Step 3 — Correction (equal weights, n = 4): c = −misclosure / n = −8'' / 4 = −2'' per angle Step 4 — Apply corrections: α₁ (adjusted) = 88°00'05'' − 2'' = 88°00'03'' α₂ (adjusted) = 92°00'03'' − 2'' = 92°00'01'' α₃ (adjusted) = 88°59'58'' − 2'' = 88°59'56'' α₄ (adjusted) = 91°00'02'' − 2'' = 91°00'00'' Step 5 — Verification: 88°00'03'' + 92°00'01'' + 88°59'56'' + 91°00'00'' Degrees: 88+92+88+91 = 359° Minutes: 00+00+59+00 = 59' Seconds: 03+01+56+00 = 60'' = 1'00'' Sum = 359°+59'+60'' = 359°+60' = 360°00'00'' ✓ Conclusion: All four angles are adjusted by −2'' each. The adjusted angles satisfy the horizon condition exactly.

Question Type

numerical

Answer Structure

• Step 1: Correct DMS summation of four angles = 360°00'08'' [1 mark]
• Step 2: Misclosure = +8'' [0.5 mark]
• Step 3: c = −2'' per angle (n = 4) [1 mark]
• Step 4: All four adjusted angles correctly stated [1.5 marks]
• Step 5: Complete verification check [1 mark]

Scoring Breakdown

Common Mark Deductions

• DMS carry-over error when summing four angles.
• Using n = 3 instead of n = 4.
• Omitting the verification step.
• Applying +2'' instead of −2''.

Key Phrases To Include

• horizon condition
• Σ = 360°00'08''
• misclosure = +8''
• c = −2'' per angle
• n = 4
• CHECK: 360°00'00'' ✓

Marks

2

Topic

Verification Step

Difficulty

easy

Template Id

T14

Examiner Tip

Use the phrase 'condition equation satisfied exactly' — this is the precise engineering language that examiners reward.

Model Answer

The verification step confirms two things: 1. Mathematical correctness: It demonstrates that the adjusted observations now satisfy the condition equation exactly — i.e., the misclosure has been fully absorbed and the required total (e.g., 180°, 360°, or zero loop closure) is achieved. If the check fails, an arithmetic error was made in applying the corrections. 2. Professional integrity: In geodetic practice and in board examinations, the verification step is the standard by which the quality of the adjustment is judged. An adjustment without verification is considered incomplete. In exam context, the verification line 'CHECK: Σadjusted = required total ✓' is often worth a dedicated mark and should never be omitted.

Question Type

short_answer

Answer Structure

• Point 1: Confirms misclosure is fully absorbed and required total is achieved [1 mark]
• Point 2: Signals completeness of the adjustment and is a distinct mark-earning step [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying 'verification checks if the answer is correct' — too vague, examiners want 'condition equation satisfied exactly'.
• Not mentioning what a failed verification implies.

Key Phrases To Include

• misclosure fully absorbed
• condition equation satisfied exactly
• required total achieved
• completeness of adjustment
• arithmetic error detected

Marks

5

Topic

Triangle Condition — Negative Misclosure

Difficulty

hard

Template Id

T15

Examiner Tip

The double-negative is a classic exam trap: misclosure = −9'', correction = −(−9'')/3 = +3''. Show the algebra explicitly — do not just write +3'' without showing the sign handling.

Model Answer

Given: Σ(observed angles) = 179°59'51'' Condition: Plane triangle — required total = 180°00'00'' Part (a) — Misclosure: Misclosure = Σ(observed) − required total Misclosure = 179°59'51'' − 180°00'00'' Convert: 179°59'51'' = 179°59'51'' and 180°00'00'' = 180°00'00'' Subtract: 179°59'51'' − 180°00'00'' = −9'' Misclosure = −9'' Part (b) — Correction per angle (equal weights, n = 3): c = −misclosure / n = −(−9'') / 3 = +9'' / 3 = +3'' per angle The correction is +3'' — each angle must be INCREASED by 3''. Part (c) — Sign convention justification: The correction is opposite in sign to the misclosure by design. Since the observed sum is TOO SMALL (negative misclosure of −9''), the angles must be increased (positive correction of +3'') to bring their sum up to the required 180°. This is the fundamental rule: correction = −misclosure / n, ensuring the correction always opposes and eliminates the misclosure. Verification: Adjusted sum = 179°59'51'' + 3(+3'') = 179°59'51'' + 9'' = 180°00'00'' ✓

Question Type

numerical

Answer Structure

• Part (a): Correct computation of misclosure = −9'' [1 mark]
• Part (b): Correct correction c = +3'' per angle with formula shown [1.5 marks]
• Part (c): Correct sign convention justification — correction opposes misclosure [1.5 marks]
• Verification check [1 mark]

Scoring Breakdown

Common Mark Deductions

• Computing misclosure as +9'' instead of −9'' (reversed subtraction).
• Applying c = −3'' (taking the correction in the wrong direction) — angles already too small would be made even smaller.
• Omitting the sign convention justification — part (c) is worth dedicated marks.
• Forgetting that two negatives make a positive: −(−9'') = +9''.

Key Phrases To Include

• misclosure = −9''
• c = −(−9'')/3 = +3''
• correction opposes misclosure
• angles must be increased
• c = −misclosure/n
• 180°00'00'' ✓