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Class 10 Mathematics Unit Wise Class Test 1 | Chapter 7 Trigonometry | Punjab Board 2027 Model Papers
Class 10 Math Chapter 7 Test 1 Trigonometry
Class 10 Mathematics Unit Wise Class Test 1 – Chapter 7: Trigonometry is designed according to the Punjab Board 2027 Model Papers and the latest board-style preparation requirements. This Class 10 Trigonometry test includes objective and subjective questions covering important concepts, trigonometric ratios, bearings, angles, applications of trigonometry, and problem-solving practice. Students can use this Punjab Board 2027 Mathematics Model Paper for chapter-wise test preparation, revision, and exam practice.
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Unit Wise Class Test
T-1
Unit 7: Trigonometry
Time: 5 minutes
(Objective Type)
Total Marks: 5
1. Tick the correct answer. (1 × 5 = 5)
(i) sin 150° =
Answer: (b) 12
Step 1: 150° lies in the second quadrant (between 90° and 180°), where sine is positive.
Step 2: Write 150° = 180° − 30°.
Step 3: sin(180° − θ) = sin θ, so sin 150° = sin 30°.
Step 4: sin 30° = 12. So sin 150° = 12.
(ii) Area of a triangle = √[s(s − a)(s − b)(s − c)] where s =
Answer: (b) a + b + c2
Step 1: This is Hero’s (Heron’s) formula for the area of a triangle with sides a, b and c.
Step 2: In it, s is the semi-perimeter, i.e. half of the perimeter.
Step 3: Perimeter = a + b + c, so s = a + b + c2.
(iii) Bearing of the point P is:
Answer: (c) 140°
Step 1: Bearing is measured clockwise from North and is written with 3 digits.
Step 2: From the diagram, P is 40° from the South line, towards the East (right side).
Step 3: From North to South is 180°, and P is 40° before reaching South (going clockwise from the East side).
Step 4: Bearing of P = 180° − 40° = 140°.
Check: 140° is between 90° (East) and 180° (South), i.e. south-east, which matches the diagram ✓
(iv) cot (−θ) =
Answer: (b) − cot θ
Step 1: cot(−θ) = cos(−θ)sin(−θ).
Step 2: cos is an even function: cos(−θ) = cos θ. sin is an odd function: sin(−θ) = −sin θ.
Step 3: cot(−θ) = cos θ−sin θ = −cos θsin θ = − cot θ.
(v) All trigonometric ratios are positive in:
Answer: (a) First quadrant
Step 1: Use the “ASTC” rule: All, Sin, Tan, Cos for quadrants I, II, III, IV.
Step 2: Quadrant I: all ratios positive. Quadrant II: only sin (and cosec). Quadrant III: only tan (and cot). Quadrant IV: only cos (and sec).
Step 3: So all ratios are positive in the first quadrant.
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Unit Wise Class Test
T-1
Unit 7: Trigonometry
Time: 35 minutes
(Subjective Type)
Total Marks: 20
2. Write short answers to the following questions: (6 × 2 = 12)
(i) A plane takes off at an angle of elevation of 12° and travels 2000 metres along this path. How high is the plane after traveling this distance?
Answer: Height ≈ 415.8 m
Step 1: The path (2000 m) is the hypotenuse of a right triangle. The height is the side opposite the 12° angle.
Step 2: sin 12° = opposite ÷ hypotenuse = h ÷ 2000.
Step 3: h = 2000 × sin 12° = 2000 × 0.2079 (sin 12° ≈ 0.2079).
Step 4: h ≈ 415.8 m (about 416 m).
(ii) In a sound lab, two microphones are placed c = 3m apart and a sound source is a = 2m from one and b = 2.5m from the other. Find the angle between the sound paths a and b.
Answer: ≈ 82.8°
Step 1: The sound source and the two microphones form a triangle with sides a = 2, b = 2.5 and c = 3. The angle γ between a and b is opposite to side c.
Step 2: Cosine rule: c² = a² + b² − 2ab cos γ.
Step 3: 3² = 2² + 2.5² − 2(2)(2.5) cos γ, so 9 = 4 + 6.25 − 10 cos γ.
Step 4: 9 = 10.25 − 10 cos γ, so 10 cos γ = 1.25, so cos γ = 0.125.
Step 5: γ = cos⁻¹(0.125) ≈ 82.8°.
(iii) Abdul Hadi walks 100m North and then 300m East. How far is he from his starting position?
Answer: ≈ 316.2 m (100√10 m)
Step 1: North and East are at right angles, so the path and the straight line back to the start form a right triangle with legs 100 m and 300 m.
Step 2: Pythagoras: d² = 100² + 300² = 10 000 + 90 000 = 100 000.
Step 3: d = √100 000 = 100√10 ≈ 316.2 m.
(iv) Fatima walked South for 5.5km and then turned West for 1.3km. Calculate Huria’s bearing from her starting point.
Answer: Bearing ≈ 193.3° (about 193°)
Step 1: South then West: she ends up in the south-west of her starting point. The path forms a right triangle: 5.5 km (South) and 1.3 km (West).
Step 2: Let α be the angle at the start between the South line and the line to her final position. tan α = opposite ÷ adjacent = 1.3 ÷ 5.5 ≈ 0.2364.
Step 3: α = tan⁻¹(0.2364) ≈ 13.3°.
Step 4: Bearing is measured clockwise from North. South is 180°, and she is 13.3° beyond South towards West.
Step 5: Bearing = 180° + 13.3° = 193.3° (≈ 193°).
Extra: Straight-line distance = √(5.5² + 1.3²) = √31.94 ≈ 5.65 km.
Note: the question says “Fatima” and then “Huria”. They are treated as the same person.
(v) A listener hears a sound from two speakers. One is 6m directly ahead and the other is at 30° to the side, 6m away. Find the distance between the speakers.
Answer: ≈ 3.11 m
Step 1: The listener and the two speakers form a triangle with two sides 6 m and 6 m and the included angle 30°.
Step 2: Cosine rule: d² = 6² + 6² − 2(6)(6) cos 30°.
Step 3: d² = 36 + 36 − 72 × 0.8660 = 72 − 62.35 = 9.65.
Step 4: d = √9.65 ≈ 3.11 m.
Check: It is an isosceles triangle, so d = 2 × 6 × sin 15° = 12 × 0.2588 ≈ 3.11 m ✓
(vi) A boat crosses a river 80m wide, at a 60° angle to the current. What distance does the boat actually travel?
Answer: ≈ 92.4 m
Step 1: The river is 80 m wide, so the straight-across distance (perpendicular to the banks) is 80 m. The boat’s path makes 60° with the current (the bank direction).
Step 2: In the right triangle, 80 m is the side opposite the 60° angle and the boat’s path is the hypotenuse d.
Step 3: sin 60° = 80 ÷ d, so d = 80 ÷ sin 60° = 80 ÷ 0.8660.
Step 4: d ≈ 92.4 m (exactly 160√3 m).
Assumption: the 60° angle is measured between the boat’s actual path and the river bank (the current direction).
Note: Answer the following Questions. (4 + 4 = 8)
3. The diagram shows a pyramid. The base, PQRS, is a horizontal square of side of 10cm. The vertex, V, is vertically above the midpoint, M and VM = 12cm. Calculate the size of angle VPM.
Answer: ∠VPM ≈ 59.5°
Step 1: V is vertically above M, so triangle VMP is a right triangle with the right angle at M.
Step 2: M is the midpoint of the square base, so PM is half of the diagonal PR.
Step 3: Diagonal of the square: PR = √(10² + 10²) = √200 = 10√2 cm.
Step 4: PM = 10√2 ÷ 2 = 5√2 ≈ 7.071 cm.
Step 5: In right triangle VMP: tan(∠VPM) = opposite ÷ adjacent = VM ÷ PM = 12 ÷ 7.071 ≈ 1.697.
Step 6: ∠VPM = tan⁻¹(1.697) ≈ 59.5°.
Check: VP = √(12² + 7.071²) = √194 ≈ 13.93 cm and sin(∠VPM) = 12 ÷ 13.93 ≈ 0.861, giving 59.5° ✓
4. In an isometric game, the camera is placed at a 45° angle from the ground. If the player is 10 units in front and 10 units above, what is the direct line of sight distance?
Answer: 10√2 ≈ 14.14 units
Step 1: The horizontal distance (10 units) and vertical height (10 units) meet at a right angle, so the line of sight is the hypotenuse.
Step 2 (Pythagoras): d² = 10² + 10² = 100 + 100 = 200.
Step 3: d = √200 = 10√2 ≈ 14.14 units.
Check with trigonometry: sin 45° = 10 ÷ d, so d = 10 ÷ 0.7071 ≈ 14.14 ✓
