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Ratios and Graphs of Trigonometric Functions · Form 5

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

Puts two or three ideas together: angles of elevation with an observer's height, solving a trigonometric equation over 0°–360°, and reading key features from a trig graph. For students ready to move past single-step questions.

Written by the spmmath.com.my editorial team.

Worked example 1

A surveyor stands on level ground 30 m from the foot of a vertical building. The angle of elevation of the top of the building from her eye is 42°.

Her eye is 1.5 m above the ground. Find the height of the building, correct to the nearest 0.1 m.

  1. Let h₁ be the height of the building above eye level. The horizontal distance is 30 m.
  2. tan 42° = h₁ / 30, so h₁ = 30 × tan 42° = 30 × 0.9004 = 27.01 m.
  3. The eye is 1.5 m above the ground, so total height = 27.01 + 1.5.
  4. Height of building = 28.51 ≈ 28.5 m.

Worked example 2

Solve the equation 2 cos x + 1 = 0 for 0° ≤ x ≤ 360°.

  1. Rearrange: 2 cos x = −1, so cos x = −1/2 = −0.5.
  2. Reference angle = cos⁻¹(0.5) = 60°.
  3. cos x is negative, so x is in the second and third quadrants.
  4. Second quadrant: x = 180° − 60° = 120°. Third quadrant: x = 180° + 60° = 240°.

Worked example 3

The graph of y = 3 sin x is drawn for 0° ≤ x ≤ 360°. (a) State the maximum and minimum values of y.

(b) State the coordinates of the maximum point. (c) State the number of solutions of the equation 3 sin x = 2 in this range.

  1. (a) The amplitude is 3, so y ranges from −3 to 3: maximum = 3, minimum = −3.
  2. (b) For y = 3 sin x, the maximum occurs when sin x = 1, i.e. x = 90°. Maximum point = (90°, 3).
  3. (c) 3 sin x = 2 means sin x = 2/3 ≈ 0.667, the horizontal line y = 2 which lies between −3 and 3.
  4. The line y = 2 cuts the single hump of the sine curve twice (once rising, once falling), so there are 2 solutions.

Worked example 4

Solve the equation 4 sin x + 3 = 1 for 0° ≤ x ≤ 360°.

  1. Rearrange: 4 sin x = 1 − 3 = −2, so sin x = −½.
  2. The basic (reference) angle is sin⁻¹(½) = 30°.
  3. Since sin x is negative, x lies in the third and fourth quadrants.
  4. Third quadrant: x = 180° + 30° = 210°. Fourth quadrant: x = 360° − 30° = 330°.

Worked example 5

From the top of a vertical cliff 80 m high, the angle of depression of a boat at sea is 35°. Calculate the horizontal distance of the boat from the foot of the cliff.

  1. The angle of depression from the top equals the angle of elevation from the boat, so the angle at the boat is 35°.
  2. The cliff (80 m) is opposite this angle and the horizontal distance d is adjacent, so tan 35° = 80 / d.
  3. Rearrange: d = 80 / tan 35°.
  4. d = 80 / 0.7002 = 114.25 m.

Worked example 6

Given that sin θ = 3/5 and θ is obtuse (90° < θ < 180°), find the values of cos θ and tan θ.

  1. For the reference angle, use a right triangle with opposite 3 and hypotenuse 5; the adjacent side is √(5² − 3²) = √16 = 4.
  2. So the reference-angle ratios are cos = 4/5 and tan = 3/4.
  3. θ is obtuse, so it lies in the second quadrant, where cosine and tangent are negative (only sine is positive).
  4. Therefore cos θ = −4/5 and tan θ = −3/4.

Worked example 7

A ladder 6.5 m long leans against a vertical wall with its foot 2.5 m from the base of the wall. Find, correct to the nearest degree, the angle the ladder makes with the ground, and find the height it reaches up the wall.

  1. Let θ be the angle between the ladder and the ground. cos θ = adjacent/hypotenuse = 2.5/6.5.
  2. θ = cos⁻¹(2.5/6.5) = cos⁻¹(0.3846) ≈ 67.4°, so θ ≈ 67° (nearest degree).
  3. Height up wall = √(6.5² − 2.5²) = √(42.25 − 6.25) = √36 = 6 m.

Worked example 8

Solve the equation 3 tan x − 5 = 1 for 0° ≤ x ≤ 360°.

  1. 3 tan x = 6, so tan x = 2.
  2. Basic/reference angle = tan⁻¹(2) = 63.4°.
  3. tan x is positive in the 1st and 3rd quadrants, so x = 63.4° or x = 180° + 63.4° = 243.4°.

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

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Frequently asked questions

What extra skill does the medium level add to basic trig ratios?

Medium questions usually move beyond one right-angled triangle, you might need to sketch a trigonometric graph over a given range, read values of angle of elevation or depression from a diagram, or use special angles (30°, 45°, 60°) without a calculator. Recognising which quadrant an angle sits in also becomes important.

What do examiners want to see in graph-based trig answers?

A neat curve at the correct scale, passing through the right key points (maximum, minimum, and where it crosses the axis), plotted over the exact given domain. For elevation/depression problems, a clearly labelled diagram showing the horizontal line and the angle is expected before any calculation.

What trips students up at medium level?

Confusing angle of elevation with angle of depression (they're measured from different horizontal lines), sketching a trig graph with the wrong amplitude or period, and forgetting that values in the second, third or fourth quadrant can be negative. Skipping the diagram for word problems is another frequent cause of lost marks.

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