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Ratios and Graphs of Trigonometric Functions · Form 5

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

Real situations where the trigonometry is hidden inside the story: you must draw the diagram, set up equations from two triangles or a motion model, and carry values through several steps. For students aiming at higher-order, application questions.

Written by the spmmath.com.my editorial team.

Worked example 1

From point A on level ground, the angle of elevation of the top of a flagpole is 34°. Walking 25 m directly towards the pole to point B, the angle of elevation becomes 48°.

Find the height of the flagpole, correct to the nearest 0.1 m.

  1. Let h be the height of the pole and x the horizontal distance from B to its foot.
  2. From B: tan 48° = h / x, so x = h / tan 48°.
  3. From A (which is 25 m further back): tan 34° = h / (x + 25), so x + 25 = h / tan 34°.
  4. Subtract the first from the second: (h / tan 34°) − (h / tan 48°) = 25.
  5. h (1/tan 34° − 1/tan 48°) = 25 → h (1.4826 − 0.9004) = 25 → h (0.5822) = 25.
  6. h = 25 / 0.5822 = 42.9 m.

Worked example 2

At 9 a.m. the sun's angle of elevation is 58° and a vertical lamp post casts a shadow 4.2 m long on level ground.

By the afternoon the shadow has lengthened to 7.5 m. Find (a) the height of the lamp post and (b) the sun's angle of elevation in the afternoon, correct to the nearest degree.

  1. (a) The post, its shadow and the sun's ray form a right triangle: tan 58° = height / 4.2.
  2. Height = 4.2 × tan 58° = 4.2 × 1.6003 = 6.72 m.
  3. (b) The post keeps the same height. In the afternoon: tan θ = 6.72 / 7.5 = 0.8962.
  4. θ = tan⁻¹(0.8962) = 41.9° ≈ 42°.

Worked example 3

A seat on a rotating wheel starts at its lowest point. As the wheel turns through an angle x, the height of the seat above its lowest point, in metres, is modelled by h = 5 − 5 cos x for 0° ≤ x ≤ 360°.

Find the angles at which the seat is 7.5 m above its lowest point.

  1. Set the model equal to 7.5: 5 − 5 cos x = 7.5.
  2. −5 cos x = 7.5 − 5 = 2.5, so cos x = −0.5.
  3. Reference angle = cos⁻¹(0.5) = 60°; cos x is negative, so x is in the second and third quadrants.
  4. x = 180° − 60° = 120° and x = 180° + 60° = 240°.

Worked example 4

A drone hovers directly above a point on a straight, level road between two observers A and B, who are 200 m apart. From A the angle of elevation of the drone is 40°, and from B it is 55°.

Calculate the height of the drone above the road, correct to 2 decimal places.

  1. Let the height be h m and let the drone be x m (horizontally) from A, so it is (200 − x) m from B.
  2. From A: tan 40° = h/x, so x = h/tan 40°. From B: tan 55° = h/(200 − x), so 200 − x = h/tan 55°.
  3. Add the two horizontal distances: h/tan 40° + h/tan 55° = 200.
  4. h(1.1918 + 0.7002) = 200 → h(1.8920) = 200 → h = 200 / 1.8920 = 105.71 m.

Worked example 5

A windmill blade of length 6 m turns about a hub fixed 8 m above the ground. As it rotates, the height of the blade tip above the ground is modelled by h = 8 + 6 sin x, where x is the angle turned, 0° ≤ x ≤ 360°.

(a) State the maximum height of the tip. (b) Find all values of x for which the tip is 5 m above the ground.

  1. (a) The maximum of sin x is 1, so the maximum height is 8 + 6(1) = 14 m.
  2. (b) Set h = 5: 8 + 6 sin x = 5, so 6 sin x = −3 and sin x = −½.
  3. The basic angle is 30°; sin x is negative, so x is in the third and fourth quadrants.
  4. x = 180° + 30° = 210° or x = 360° − 30° = 330°.

Worked example 6

A vertical tower stands on level ground with a flag pole mounted vertically on top of it. From a point P on the ground 18 m from the foot of the tower, the angle of elevation of the top of the tower is 52° and the angle of elevation of the top of the flag pole is 58°.

Calculate the height of the flag pole, correct to 2 decimal places.

  1. Height of the top of the tower above the ground: h₁ = 18 tan 52° = 18 × 1.2799 = 23.04 m.
  2. Height of the top of the flag pole above the ground: h₂ = 18 tan 58° = 18 × 1.6003 = 28.81 m.
  3. The flag pole is the part between these two heights: flag pole = h₂ − h₁.
  4. Flag pole = 28.81 − 23.04 = 5.77 m.

Worked example 7

A tall pole and a short pole (4 m high) stand on the same level ground, the short pole nearer to an observer. From the foot of the short pole, the angle of elevation of the top of the tall pole is 40°.

From the top of the short pole, the angle of elevation of the top of the tall pole is 32°. Find the height of the tall pole, correct to 2 decimal places.

  1. Let H = height of the tall pole and d = horizontal distance between the poles.
  2. From the foot of the short pole: tan 40° = H/d, so H = d tan 40°.
  3. From the top of the short pole (height 4 m): tan 32° = (H − 4)/d, so H = d tan 32° + 4.
  4. Equate: d tan 40° = d tan 32° + 4 → d(tan 40° − tan 32°) = 4 → d = 4/(0.8391 − 0.6249) = 4/0.2142 ≈ 18.67 m.
  5. H = d tan 40° = 18.67 × 0.8391 ≈ 15.67 m.

Worked example 8

Standing at point A on one bank of a straight river, directly opposite a tree T on the far bank, a surveyor walks 50 m along the bank to point B. The angle TBA is found to be 34°.

(a) Calculate the width of the river. (b) From B, the angle of elevation of the top of the tree is 8°.

Find the height of the tree, correct to 2 decimal places.

  1. In right triangle TAB (right angle at A): tan 34° = AT/AB = AT/50, so AT = 50 tan 34° = 50 × 0.6745 ≈ 33.73 m, this is the width of the river.
  2. Find the ground distance BT: cos 34° = AB/BT = 50/BT, so BT = 50/cos 34° = 50/0.8290 ≈ 60.31 m.
  3. In the vertical triangle at B: tan 8° = height of tree/BT, so height = 60.31 × tan 8° = 60.31 × 0.1405 ≈ 8.48 m.

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

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Frequently asked questions

What makes KBAT trigonometry questions harder than the graph or ratio questions before them?

KBAT items typically embed trigonometry in a real or unfamiliar context, two connected triangles sharing a side, a problem solved by reading an intersection off a sketched graph, or a situation requiring you to decide which trig ratio applies before any numbers are given. You must plan the whole solution path first.

What does the examiner reward in a KBAT trigonometry solution?

Marks go to a logically ordered chain of working, identifying the correct triangle or graph feature, stating the ratio or reading used, and connecting each step to the next with a reason. A correct final answer with no visible reasoning earns far less than an incomplete but well-structured attempt.

What pitfalls catch strong students in this KBAT set?

Rushing into a calculation before sketching the situation, so the wrong angle or side gets used; treating a two-triangle problem as one triangle; and, for graph-based questions, misreading the scale on the axes when estimating an intersection point. Slowing down to plan usually prevents all three.

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