USTET Mathematics — TrigonometryExam Answer Templates

Marks

1

Topic

Special Angles

Difficulty

easy

Template Id

T1

Examiner Tip

Always use exact fractional values for special angles unless specifically asked for decimal approximation

Model Answer

sin 30° = 1/2

Question Type

very_short_answer

Answer Structure

• Line 1: Direct value from special angle table [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing decimal 0.5 instead of fraction 1/2
• Incorrect value

Key Phrases To Include

• 1/2
• exact value

Marks

2

Topic

Angle Conversion

Difficulty

easy

Template Id

T2

Examiner Tip

Always write the conversion formula first to show understanding of the method

Model Answer

Given: Angle = 45° To convert to radians: Using the formula: radians = degrees × π/180° = 45° × π/180° = π/4 radians Therefore, 45° = π/4 radians

Question Type

short_answer

Answer Structure

• Line 1: State the conversion formula [1 mark]
• Line 2: Substitute and calculate correctly [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not showing the conversion formula
• Incorrect calculation
• Missing π in the answer

Key Phrases To Include

• π/180°
• π/4 radians
• conversion formula

Marks

3

Topic

Trigonometric Ratios

Difficulty

medium

Template Id

T3

Examiner Tip

Always find the missing side first using Pythagorean theorem, then apply SOH-CAH-TOA systematically

Model Answer

Given: Opposite side = 3 units, Hypotenuse = 5 units To find: sin θ, cos θ, tan θ First, find adjacent side using Pythagorean theorem: adjacent² + opposite² = hypotenuse² adjacent² + 3² = 5² adjacent² = 25 - 9 = 16 adjacent = 4 units Now: sin θ = opposite/hypotenuse = 3/5 cos θ = adjacent/hypotenuse = 4/5 tan θ = opposite/adjacent = 3/4 Therefore: sin θ = 3/5, cos θ = 4/5, tan θ = 3/4

Question Type

short_answer

Answer Structure

• Line 1-3: Find missing side using Pythagorean theorem [1 mark]
• Line 4: Calculate sin θ correctly [1 mark]
• Line 5-6: Calculate cos θ and tan θ correctly [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not finding the adjacent side first
• Incorrect trigonometric ratios
• Arithmetic errors

Key Phrases To Include

• Pythagorean theorem
• opposite/hypotenuse
• adjacent/hypotenuse
• opposite/adjacent

Marks

3

Topic

Trigonometric Identities

Difficulty

medium

Template Id

T4

Examiner Tip

Always consider the quadrant to determine the correct sign of trigonometric functions

Model Answer

Given: sin θ = 3/5, θ is in first quadrant To find: cos θ and tan θ Using the identity: sin²θ + cos²θ = 1 (3/5)² + cos²θ = 1 9/25 + cos²θ = 1 cos²θ = 1 - 9/25 = 16/25 cos θ = ±4/5 Since θ is in first quadrant, cos θ > 0 Therefore, cos θ = 4/5 tan θ = sin θ/cos θ = (3/5)/(4/5) = 3/4 Therefore: cos θ = 4/5, tan θ = 3/4

Question Type

short_answer

Answer Structure

• Line 1-4: Use Pythagorean identity to find cos²θ [1 mark]
• Line 5-6: Determine correct sign based on quadrant [1 mark]
• Line 7: Calculate tan θ using ratio identity [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not considering quadrant signs
• Arithmetic errors
• Not using correct trigonometric identity

Key Phrases To Include

• Pythagorean identity
• first quadrant
• sin²θ + cos²θ = 1
• tan θ = sin θ/cos θ

Marks

3

Topic

Trigonometric Identities

Difficulty

medium

Template Id

T5

Examiner Tip

Always start proofs with a fundamental identity and show each algebraic step clearly

Model Answer

To prove: tan²θ + 1 = sec²θ Starting with the fundamental identity: sin²θ + cos²θ = 1 Dividing both sides by cos²θ: (sin²θ + cos²θ)/cos²θ = 1/cos²θ sin²θ/cos²θ + cos²θ/cos²θ = 1/cos²θ (sin θ/cos θ)² + 1 = (1/cos θ)² tan²θ + 1 = sec²θ Hence proved.

Question Type

short_answer

Answer Structure

• Line 1: Start with fundamental Pythagorean identity [1 mark]
• Line 2-4: Divide by cos²θ and simplify correctly [1 mark]
• Line 5: Recognize tan θ and sec θ definitions [1 mark]

Scoring Breakdown

Common Mark Deductions

• Starting with wrong identity
• Algebraic errors
• Not showing clear logical steps

Key Phrases To Include

• fundamental identity
• dividing by cos²θ
• tan θ = sin θ/cos θ
• sec θ = 1/cos θ
• Hence proved

Marks

3

Topic

Applications of Trigonometry

Difficulty

medium

Template Id

T6

Examiner Tip

Always draw a clear diagram for word problems and use exact values for special angles

Model Answer

Given: Length of ladder = 10 m, Angle with ground = 60° To find: Height at which ladder touches the wall From the right triangle formed: - Hypotenuse = ladder length = 10 m - Angle between ladder and ground = 60° - Height = opposite side to the 60° angle Using sin 60° = opposite/hypotenuse sin 60° = height/10 √3/2 = height/10 height = 10 × √3/2 = 5√3 m Therefore, the ladder touches the wall at a height of 5√3 meters.

Question Type

short_answer

Answer Structure

• Line 1-2: Identify the right triangle and given values [1 mark]
• Line 3-4: Choose correct trigonometric ratio (sin 60°) [1 mark]
• Line 5-6: Calculate correctly using exact values [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using decimal values instead of exact values
• Wrong trigonometric ratio
• Not drawing or visualizing the triangle correctly

Key Phrases To Include

• right triangle
• sin 60° = √3/2
• opposite side
• 5√3 meters

Marks

2

Topic

Reference Angles

Difficulty

easy

Template Id

T7

Examiner Tip

Remember the reference angle formulas for each quadrant: Q2: 180°-θ, Q3: θ-180°, Q4: 360°-θ

Model Answer

Given: Angle = 150° To find: Reference angle 150° lies in the second quadrant (90° < 150° < 180°) For angles in the second quadrant, reference angle = 180° - angle Reference angle = 180° - 150° = 30° Therefore, the reference angle for 150° is 30°.

Question Type

short_answer

Answer Structure

• Line 1: Identify which quadrant the angle is in [1 mark]
• Line 2: Apply correct formula and calculate [1 mark]

Scoring Breakdown

Common Mark Deductions

• Wrong quadrant identification
• Using wrong formula for reference angle

Key Phrases To Include

• second quadrant
• 180° - angle
• reference angle = 30°

Marks

2

Topic

Trigonometric Functions in Different Quadrants

Difficulty

medium

Template Id

T8

Examiner Tip

Remember: in quadrant II, only sine and cosecant are positive

Model Answer

Given: Find sin 150° 150° lies in the second quadrant Reference angle = 180° - 150° = 30° In second quadrant, sine is positive sin 150° = sin 30° = 1/2 Therefore, sin 150° = 1/2

Question Type

short_answer

Answer Structure

• Line 1-2: Find reference angle [1 mark]
• Line 3-4: Apply quadrant rule and calculate [1 mark]

Scoring Breakdown

Common Mark Deductions

• Wrong sign for the quadrant
• Incorrect reference angle calculation

Key Phrases To Include

• second quadrant
• reference angle = 30°
• sine is positive
• 1/2

Marks

3

Topic

Quadrant Analysis

Difficulty

medium

Template Id

T9

Examiner Tip

Use the signs of trigonometric functions to identify quadrants: Q2 has sin > 0, cos < 0

Model Answer

Given: cos θ = -3/5, sin θ > 0 To find: Quadrant and tan θ Since cos θ < 0 and sin θ > 0, θ is in the second quadrant. Using sin²θ + cos²θ = 1: sin²θ + (-3/5)² = 1 sin²θ + 9/25 = 1 sin²θ = 16/25 sin θ = ±4/5 Since sin θ > 0, we have sin θ = 4/5 tan θ = sin θ/cos θ = (4/5)/(-3/5) = -4/3 Therefore: θ is in quadrant II, and tan θ = -4/3

Question Type

short_answer

Answer Structure

• Line 1: Identify quadrant using signs [1 mark]
• Line 2-4: Find sin θ using Pythagorean identity [1 mark]
• Line 5: Calculate tan θ correctly [1 mark]

Scoring Breakdown

Common Mark Deductions

• Wrong quadrant identification
• Sign errors in calculations
• Not using the condition sin θ > 0

Key Phrases To Include

• second quadrant
• Pythagorean identity
• sin θ = 4/5
• tan θ = -4/3

Marks

3

Topic

Trigonometric Equations

Difficulty

medium

Template Id

T10

Examiner Tip

Always find all solutions within the given range by considering all quadrants where the function is positive

Model Answer

Given equation: 2sin θ - 1 = 0 Range: 0° ≤ θ ≤ 360° 2sin θ = 1 sin θ = 1/2 The reference angle where sin θ = 1/2 is 30° Sine is positive in quadrants I and II In quadrant I: θ = 30° In quadrant II: θ = 180° - 30° = 150° Therefore, θ = 30° or θ = 150°

Question Type

short_answer

Answer Structure

• Line 1-2: Solve for sin θ [1 mark]
• Line 3: Identify reference angle [1 mark]
• Line 4-5: Find all solutions in given range [1 mark]

Scoring Breakdown

Common Mark Deductions

• Missing one solution
• Not considering the given range
• Incorrect reference angle

Key Phrases To Include

• sin θ = 1/2
• reference angle = 30°
• quadrants I and II
• θ = 30° or 150°

Marks

5

Topic

Law of Sines

Difficulty

hard

Template Id

T11

Examiner Tip

Always use exact trigonometric values for special angles and show each algebraic step clearly

Model Answer

Given: Triangle ABC with A = 60°, B = 45°, b = 8 units To find: Side 'a' Step 1: State the Law of Sines a/sin A = b/sin B = c/sin C Step 2: Substitute known values a/sin 60° = 8/sin 45° Step 3: Find the values of trigonometric functions sin 60° = √3/2 sin 45° = √2/2 Step 4: Substitute and solve for 'a' a/(√3/2) = 8/(√2/2) a × (2/√3) = 8 × (2/√2) a × (2/√3) = 16/√2 Step 5: Simplify a = (16/√2) × (√3/2) a = 16√3/(2√2) a = 8√3/√2 a = 8√3/√2 × √2/√2 a = 8√6/2 a = 4√6 units Therefore, side a = 4√6 units ≈ 9.80 units

Question Type

long_answer

Answer Structure

• Line 1: State the Law of Sines formula [1 mark]
• Line 2: Substitute given values correctly [1 mark]
• Line 3: Write exact values of trig functions [1 mark]
• Line 4-5: Algebraic manipulation and solving [1 mark]
• Line 6: Final answer in exact form [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using decimal approximations instead of exact values
• Algebraic errors
• Not stating the law clearly
• Incomplete working

Key Phrases To Include

• Law of Sines
• exact values
• sin 60° = √3/2
• sin 45° = √2/2
• 4√6 units

Marks

4

Topic

Applications - Angle of Elevation

Difficulty

medium

Template Id

T12

Examiner Tip

For angle of elevation problems, always use tangent ratio when you have the horizontal distance and need height

Model Answer

Given: Shadow length = 20 m, Angle of elevation = 35° To find: Height of the tree Step 1: Draw and analyze the right triangle - Base (shadow) = 20 m - Angle of elevation = 35° - Height = opposite side Step 2: Choose appropriate trigonometric ratio tan 35° = opposite/adjacent = height/shadow tan 35° = height/20 Step 3: Solve for height height = 20 × tan 35° height = 20 × 0.7002 (using calculator) height = 14.004 m Step 4: Round to nearest meter height ≈ 14 m Therefore, the height of the tree is approximately 14 meters.

Question Type

long_answer

Answer Structure

• Line 1: Set up the right triangle correctly [1 mark]
• Line 2: Choose correct trigonometric ratio [1 mark]
• Line 3: Calculate using tan 35° [1 mark]
• Line 4: Round to nearest meter as asked [1 mark]

Scoring Breakdown

Common Mark Deductions

• Wrong trigonometric ratio
• Not rounding as requested
• Calculator errors
• Poor diagram or setup

Key Phrases To Include

• right triangle
• angle of elevation
• tan 35°
• opposite/adjacent
• nearest meter

Marks

5

Topic

Sum and Difference Formulas

Difficulty

hard

Template Id

T13

Examiner Tip

Always determine the signs of trigonometric functions based on their quadrants before applying sum formulas

Model Answer

Given: sin A = 3/5 (A in QII), cos B = 12/13 (B in QI) To find: sin(A + B) Step 1: Find cos A using Pythagorean identity sin²A + cos²A = 1 (3/5)² + cos²A = 1 9/25 + cos²A = 1 cos²A = 16/25 cos A = ±4/5 Since A is in QII, cos A < 0, so cos A = -4/5 Step 2: Find sin B using Pythagorean identity sin²B + cos²B = 1 sin²B + (12/13)² = 1 sin²B + 144/169 = 1 sin²B = 25/169 sin B = ±5/13 Since B is in QI, sin B > 0, so sin B = 5/13 Step 3: Apply sum formula sin(A + B) = sin A cos B + cos A sin B = (3/5)(12/13) + (-4/5)(5/13) = 36/65 - 20/65 = 16/65 Therefore, sin(A + B) = 16/65

Question Type

long_answer

Answer Structure

• Line 1-3: Find cos A using identity and quadrant [2 marks]
• Line 4-5: Find sin B using identity and quadrant [1 mark]
• Line 6-7: Apply sum formula correctly [1 mark]
• Line 8: Calculate final answer [1 mark]

Scoring Breakdown

Common Mark Deductions

• Wrong signs for quadrants
• Incorrect application of sum formula
• Arithmetic errors
• Not using exact fractional form

Key Phrases To Include

• Pythagorean identity
• quadrant signs
• sum formula
• sin(A + B) = sin A cos B + cos A sin B
• 16/65

Marks

4

Topic

Trigonometric Equations with Multiple Angles

Difficulty

hard

Template Id

T14

Examiner Tip

For equations with multiple angles like 2θ, remember to find all solutions in the extended range before dividing

Model Answer

Given: cos 2θ = 1/2 Range: 0° ≤ θ ≤ 360° Step 1: Solve for 2θ cos 2θ = 1/2 The reference angle where cos = 1/2 is 60° Cosine is positive in quadrants I and IV Step 2: Find all values of 2θ in [0°, 720°] In QI: 2θ = 60° In QIV: 2θ = 360° - 60° = 300° For one complete cycle: 2θ = 60°, 300° For second cycle: 2θ = 60° + 360° = 420°, 300° + 360° = 660° So 2θ = 60°, 300°, 420°, 660° Step 3: Solve for θ θ = 30°, 150°, 210°, 330° Step 4: Verify all solutions are in [0°, 360°] All values 30°, 150°, 210°, 330° are within the given range. Therefore, θ = 30°, 150°, 210°, 330°

Question Type

long_answer

Answer Structure

• Line 1: Find reference angle for cos 2θ = 1/2 [1 mark]
• Line 2: Find all values of 2θ considering range [1 mark]
• Line 3: Divide by 2 to get θ values [1 mark]
• Line 4: Verify solutions are in given range [1 mark]

Scoring Breakdown

Common Mark Deductions

• Missing solutions
• Not considering the extended range for 2θ
• Incorrect division
• Not verifying final answers

Key Phrases To Include

• reference angle = 60°
• quadrants I and IV
• extended range for 2θ
• θ = 30°, 150°, 210°, 330°

Marks

2

Topic

Domain and Range of Trigonometric Functions

Difficulty

medium

Template Id

T15

Examiner Tip

For transformed trig functions, apply transformations step by step: amplitude first, then vertical shift

Model Answer

Given: f(x) = 3sin(2x - π/4) + 1 Domain: Since sine function is defined for all real numbers, and 2x - π/4 can take any real value, the domain is all real numbers. Domain = (-∞, ∞) or ℝ Range: The sine function has range [-1, 1] Multiplying by 3: 3sin(2x - π/4) has range [-3, 3] Adding 1: f(x) has range [-3+1, 3+1] = [-2, 4] Range = [-2, 4] Therefore: Domain = ℝ, Range = [-2, 4]

Question Type

short_answer

Answer Structure

• Line 1-2: Identify domain as all real numbers [1 mark]
• Line 3-5: Find range using transformations [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not considering the transformations
• Incorrect interval notation
• Missing amplitude or shift effects

Key Phrases To Include

• all real numbers
• sine function range [-1, 1]
• amplitude 3
• vertical shift +1
• range [-2, 4]