Noesis
Level 1 — RecognitionTrigonometry — Foundation

Trigonometry — Foundation

⏱ 20 minutes30 marksprintable — key stays hidden on paper

Level 1 — Recognition Test

Time Limit: 20 minutes Total Marks: 30

Section A — Multiple Choice Questions (1 mark each)

Choose the correct option.

Q1. In a right triangle, sin⁡θ\sin\theta is defined as:

  • (a) adjacent / hypotenuse
  • (b) opposite / hypotenuse
  • (c) opposite / adjacent
  • (d) hypotenuse / opposite

Q2. The value of cos⁡60°\cos 60° is:

  • (a) 12\dfrac{1}{2}
  • (b) 32\dfrac{\sqrt{3}}{2}
  • (c) 12\dfrac{1}{\sqrt{2}}
  • (d) 11

Q3. The reciprocal of tan⁡θ\tan\theta is:

  • (a) sec⁡θ\sec\theta
  • (b) csc⁡θ\csc\theta
  • (c) cot⁡θ\cot\theta
  • (d) cos⁡θ\cos\theta

Q4. sin⁡(90°−θ)\sin(90° - \theta) equals:

  • (a) cos⁡θ\cos\theta
  • (b) sin⁡θ\sin\theta
  • (c) tan⁡θ\tan\theta
  • (d) cot⁡θ\cot\theta

Q5. The value of tan⁡45°\tan 45° is:

  • (a) 00
  • (b) 11
  • (c) 3\sqrt{3}
  • (d) undefined

Q6. By the Pythagorean theorem, in a right triangle with legs a,ba, b and hypotenuse cc:

  • (a) a+b=ca + b = c
  • (b) a2−b2=c2a^2 - b^2 = c^2
  • (c) a2+b2=c2a^2 + b^2 = c^2
  • (d) a2+b2=2c2a^2 + b^2 = 2c^2

Q7. The mnemonic "SOH" stands for:

  • (a) Sine = Opposite / Hypotenuse
  • (b) Sine = Opposite / Height
  • (c) Secant = Opposite / Hypotenuse
  • (d) Sine = Adjacent / Hypotenuse

Q8. sec⁡θ\sec\theta is defined as:

  • (a) 1sin⁡θ\dfrac{1}{\sin\theta}
  • (b) 1cos⁡θ\dfrac{1}{\cos\theta}
  • (c) 1tan⁡θ\dfrac{1}{\tan\theta}
  • (d) sin⁡θcos⁡θ\dfrac{\sin\theta}{\cos\theta}

Q9. The value of sin⁡0°\sin 0° is:

  • (a) 11
  • (b) 12\dfrac{1}{2}
  • (c) 00
  • (d) undefined

Q10. A triangle has sides 3,4,53, 4, 5. It is:

  • (a) not a right triangle
  • (b) a right triangle (by converse of Pythagoras)
  • (c) an equilateral triangle
  • (d) an obtuse triangle

Section B — Matching (1 mark each row = 4 marks)

Q11. Match Column A with the correct value in Column B.

Column A Column B
(i) sin⁡30°\sin 30° (P) 12\dfrac{1}{\sqrt{2}}
(ii) cos⁡45°\cos 45° (Q) 3\sqrt{3}
(iii) tan⁡60°\tan 60° (R) 12\dfrac{1}{2}
(iv) cos⁡0°\cos 0° (S) 11

Section C — True/False WITH Justification (2 marks each)

State True or False and give a one-line reason. (1 mark answer + 1 mark justification)

Q12. tan⁡θ=sin⁡θcos⁡θ\tan\theta = \dfrac{\sin\theta}{\cos\theta}.

Q13. cos⁡(90°−30°)=cos⁡30°\cos(90° - 30°) = \cos 30°.

Q14. csc⁡θ×sin⁡θ=1\csc\theta \times \sin\theta = 1 for any valid θ\theta.

Q15. A triangle with sides 6,8,116, 8, 11 is a right triangle.

Q16. sin⁡90°=1\sin 90° = 1.

Q17. In an angle of elevation problem, the angle is measured from the horizontal line up to the line of sight.

Section D — Short Application (3 marks)

Q18. A ladder leans against a wall making an angle of 60°60° with the ground. If the ladder is 1010 m long, how high up the wall does it reach? (Use sin⁡60°=32\sin 60° = \dfrac{\sqrt3}{2}.)

Answer keyMark scheme & solutions

Section A (10 marks)

Q1. (b) — sine = opposite/hypotenuse (SOH). [1]

Q2. (a) 12\tfrac12 — from the 30-60-90 triangle, cos⁡60°\cos 60° = adjacent/hyp = 1/21/2. [1]

Q3. (c) cot⁡θ\cot\theta — by definition cot⁡θ=1/tan⁡θ\cot\theta = 1/\tan\theta. [1]

Q4. (a) cos⁡θ\cos\theta — complementary angle relation. [1]

Q5. (b) 11 — in a 45-45-90 triangle opposite = adjacent, so tan⁡45°=1\tan 45° = 1. [1]

Q6. (c) a2+b2=c2a^2+b^2=c^2 — Pythagorean theorem. [1]

Q7. (a) Sine = Opposite / Hypotenuse. [1]

Q8. (b) 1/cos⁡θ1/\cos\theta — reciprocal identity. [1]

Q9. (c) 00 — sin⁡0°=0\sin 0° = 0. [1]

Q10. (b) right triangle — since 32+42=9+16=25=523^2+4^2 = 9+16 = 25 = 5^2 (converse of Pythagoras). [1]

Section B (4 marks)

Q11.

  • (i) → (R): sin⁡30°=12\sin 30° = \tfrac12
  • (ii) → (P): cos⁡45°=12\cos 45° = \tfrac{1}{\sqrt2}
  • (iii) → (Q): tan⁡60°=3\tan 60° = \sqrt3
  • (iv) → (S): cos⁡0°=1\cos 0° = 1

[1 each = 4]

Section C (12 marks)

Q12. True. [1] Because tan⁡θ=oppadj=opp/hypadj/hyp=sin⁡θcos⁡θ\tan\theta = \frac{\text{opp}}{\text{adj}} = \frac{\text{opp}/\text{hyp}}{\text{adj}/\text{hyp}} = \frac{\sin\theta}{\cos\theta}. [1]

Q13. False. [1] cos⁡(90°−30°)=cos⁡60°=12\cos(90°-30°) = \cos 60° = \tfrac12, which equals sin⁡30°\sin 30°, not cos⁡30°=32\cos 30°=\tfrac{\sqrt3}{2}. [1]

Q14. True. [1] Since csc⁡θ=1/sin⁡θ\csc\theta = 1/\sin\theta, the product is 1sin⁡θ⋅sin⁡θ=1\frac{1}{\sin\theta}\cdot\sin\theta = 1. [1]

Q15. False. [1] 62+82=36+64=100≠121=1126^2+8^2 = 36+64 = 100 \ne 121 = 11^2, so it fails the converse of Pythagoras. [1]

Q16. True. [1] From the unit-circle/standard-angle table, sin⁡90°=1\sin 90° = 1. [1]

Q17. True. [1] Angle of elevation is measured upward from the horizontal to the line of sight. [1]

Section D (3 marks)

Q18. Height h=ladder×sin⁡60°h = \text{ladder}\times\sin 60°. [1] h=10×32=53h = 10 \times \dfrac{\sqrt3}{2} = 5\sqrt3 m. [1] ≈8.66\approx 8.66 m. [1]

[
  {"claim":"cos60 = 1/2","code":"result = cos(rad(60)) == Rational(1,2)"},
  {"claim":"3-4-5 is right triangle","code":"result = (3**2 + 4**2 == 5**2)"},
  {"claim":"cos(90-30)=sin30 not cos30","code":"result = (cos(rad(60)) == sin(rad(30))) and (cos(rad(60)) != cos(rad(30)))"},
  {"claim":"ladder height 10*sin60 = 5*sqrt3","code":"result = simplify(10*sin(rad(60)) - 5*sqrt(3)) == 0"},
  {"claim":"6-8-11 not right triangle","code":"result = (6**2 + 8**2 != 11**2)"}
]