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Ratios and Graphs of Trigonometric Functions · Form 5

Ratios and Graphs of Trigonometric Functions: Worked Examples (Easier)

Builds the foundation: reading sine, cosine and tangent from a right-angled triangle, and finding the value of a ratio for angles up to 360° using quadrants and reference angles. Best for students starting the chapter or rebuilding basics.

Written by the spmmath.com.my editorial team.

Worked example 1

Triangle PQR is right-angled at Q, with PQ = 8 cm and QR = 6 cm. Find the values of sin P, cos P and tan P.

  1. Right angle at Q, so PR is the hypotenuse. By Pythagoras: PR = √(8² + 6²) = √(64 + 36) = √100 = 10 cm.
  2. For angle P: opposite side = QR = 6 cm, adjacent side = PQ = 8 cm, hypotenuse = PR = 10 cm.
  3. sin P = opposite ÷ hypotenuse = 6/10 = 0.6.
  4. cos P = adjacent ÷ hypotenuse = 8/10 = 0.8.
  5. tan P = opposite ÷ adjacent = 6/8 = 0.75.

Worked example 2

Without using a calculator, find the value of sin 150° using the quadrant and reference angle.

  1. 150° lies in the second quadrant (between 90° and 180°).
  2. Reference angle = 180° − 150° = 30°.
  3. In the second quadrant sine is positive (the 'All, Sine, Tangent, Cosine' quadrant rule).
  4. So sin 150° = + sin 30° = 1/2 = 0.5.

Worked example 3

Find the value of tan 210°, giving your answer correct to 3 decimal places.

  1. 210° lies in the third quadrant (between 180° and 270°).
  2. Reference angle = 210° − 180° = 30°.
  3. In the third quadrant tangent is positive (sine and cosine are both negative there, so their ratio is positive).
  4. So tan 210° = + tan 30° = 1/√3 = 0.577 (3 d.p.).

Worked example 4

In right-angled triangle ABC, the right angle is at B, AB = 12 cm and BC = 5 cm. Find the length of AC and the value of sin A.

  1. Use Pythagoras' theorem: AC² = AB² + BC² = 12² + 5² = 144 + 25 = 169.
  2. So AC = √169 = 13 cm.
  3. Angle A is at vertex A, so the side opposite A is BC = 5 and the hypotenuse is AC = 13.
  4. sin A = opposite ÷ hypotenuse = 5/13.

Worked example 5

Without using a calculator, find the value of cos 120°.

  1. 120° lies in the second quadrant, where cosine is negative.
  2. The reference angle is 180° − 120° = 60°.
  3. So cos 120° = −cos 60°.
  4. Since cos 60° = ½, cos 120° = −½.

Worked example 6

Find the value of tan 300°, giving your answer correct to 3 decimal places.

  1. 300° lies in the fourth quadrant, where tangent is negative.
  2. The reference angle is 360° − 300° = 60°.
  3. So tan 300° = −tan 60°.
  4. tan 60° = 1.7320508..., so tan 300° = −1.7320508...

Worked example 7

Triangle LMN is right-angled at M, with LM = 9 cm and MN = 12 cm. Find the length of LN and the value of cos N.

  1. Use Pythagoras' theorem: LN² = LM² + MN² = 9² + 12² = 81 + 144 = 225.
  2. LN = √225 = 15 cm.
  3. cos N = adjacent/hypotenuse = MN/LN = 12/15 = 4/5 (0.8).

Worked example 8

Without using a calculator, find the value of sin 225°.

  1. 225° lies in the third quadrant (180° < 225° < 270°), where sine is negative.
  2. The reference angle is 225° − 180° = 45°.
  3. sin 225° = −sin 45° = −√2/2 (≈ −0.707).

Source:DSKP KSSM Mathematics Form 4 and 5 (Versi English)

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Frequently asked questions

What's the best way to start an easy trigonometric-ratio question?

Sketch or label the right-angled triangle first, marking the given sides and the angle you need. Decide which ratio (sine, cosine or tangent) connects the sides you have to the one you want, using SOH-CAH-TOA, then substitute values directly.

Getting the triangle labelled correctly prevents most early mistakes.

What do examiners reward on these easier trig questions?

Correct identification of the opposite, adjacent and hypotenuse relative to the given angle, the right ratio chosen, and the calculation carried out with enough decimal places (usually 3 or 4 significant figures) before rounding the final answer. A clearly labelled diagram alone can earn a mark even if the final number is off.

What common mistakes happen at this easy level?

Mixing up which side is opposite versus adjacent when the angle is at a different vertex, forgetting to check the calculator is in degree mode, and rounding too early in a multi-step calculation. Also, some students apply the wrong ratio because they didn't first identify which two sides are involved.

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