6.3 Trigonometry Applications

From Identities to Geometry Problems

Section 6.2 built algebraic trigonometry; this section applies the same ratios to heights and distances, bearings, and aeronautically flavoured angle problems at FSc depth. These word problems appear throughout engineering entry maths banks and fit the PAF CAE style: one clear figure, one or two trig steps, integer or simple-radical answers. With ~30 seconds per maths MCQ in commonly reported papers, sketch first, then pick sin, cos, or tan.

Core Right-Triangle Ratios (SOH-CAH-TOA)

For acute angle θ in a right triangle:

Angles of Elevation and Depression

• Angle of elevation: angle between the horizontal and the line of sight up to an object.
• Angle of depression: angle between the horizontal and the line of sight down to an object.

By alternate interior angles (parallel horizontals), the depression angle from a tower top to a point on the ground equals the elevation angle from that point up to the tower top. That equality is an exam favourite.

Worked — single elevation.

From a point 40 m from the base of a tower, the angle of elevation of the top is 30°. Find the tower height h.

tan 30° = h / 40 ⇒ (1/√3) = h/40 ⇒ h = 40/√3 m (or (40√3)/3 m rationalised).

Worked — two elevations (standard “moving closer”).

The angle of elevation of a tower top from a point A on level ground is 30°. After walking 20 m toward the tower to point B, the angle becomes 60°. Find the height h.

Let the foot of the tower be C, and let BC = x. Then AC = x + 20.

From B: tan 60° = h/x ⇒ √3 = h/x ⇒ h = x√3.
From A: tan 30° = h/(x+20) ⇒ 1/√3 = h/(x+20) ⇒ h = (x+20)/√3.

Equate: x√3 = (x+20)/√3 ⇒ 3x = x + 20 ⇒ 2x = 20 ⇒ x = 10.
h = 10√3 m.

Worked — depression.

From the top of a 50 m lighthouse, the angle of depression of a boat is 45°. Horizontal distance to the boat:

tan 45° = 50 / d ⇒ 1 = 50/d ⇒ d = 50 m.

Bearings

A bearing states direction as an angle measured clockwise from north, usually written as three figures: 000° to 360° (e.g. 045°, 180°, 315°).

Older “N30°E” style appears in some textbooks: start at north, turn 30° toward east—equivalent to bearing 030°. Convert everything to a right triangle with N–S and E–W legs before computing.

Worked — bearing distance.

An aircraft flies 100 km on a bearing of 060°. How far east and how far north has it travelled? (Treat as horizontal ground track for the maths model.)

060° is 60° east of north, so:

north component = 100 cos 60° = 100 · 1/2 = 50 km,
east component = 100 sin 60° = 100 · √3/2 = 50√3 km.

(If your sketch puts the bearing angle at north, adjacent to north is cos; opposite toward east is sin—match the sketch, not a memorised “sin = east” chant without a figure.)

Worked — return bearing idea.

If you travel from P to Q on bearing θ, the reverse bearing from Q to P is θ + 180° (mod 360°). Example: outbound 040° ⇒ return 220°. Useful for “shortest path back” MCQs.

Two-Station / Observer Problems

Worked — observers on opposite sides.

Two observers A and B, 200 m apart on level ground, measure elevations 45° and 30° to the same tower top T between them. Find height h.

Let foot be C, AC = x, CB = 200 − x.

h = x tan 45° = x,
h = (200 − x) tan 30° = (200 − x)/√3.

So x = (200 − x)/√3 ⇒ x√3 = 200 − x ⇒ x(√3 + 1) = 200 ⇒ x = 200/(√3 + 1).
Rationalise: x = 200(√3 − 1)/(3 − 1) = 100(√3 − 1).
h = 100(√3 − 1) m.

Aeronautically Relevant Angles (FSc Model)

You will not need full flight-dynamics theory for the initial academic paper, but wordings often borrow aviation flavour. Map them to school geometry:

Worked — climb angle.

An aircraft gains 300 m altitude over 1500 m ground distance (constant climb). Climb angle γ satisfies

tan γ = 300/1500 = 1/5 ⇒ γ = tan⁻¹(0.2).

MCQs may ask tan γ = 1/5, or sin γ if they give slant range instead of ground run.

Worked — slant range.

A radar measures slant range 5 km to an aircraft at elevation angle 30°. Height above radar level:

h = 5 sin 30° = 5 · 1/2 = 2.5 km.

Worked — glide.

A glider descends 200 m while covering 1000 m over ground. Descent angle δ: tan δ = 200/1000 = 1/5. Glide ratio = ground/height = 5:1.

Small Checklist for Word Problems

1. Draw the tower/aircraft, observer, and horizontal.
2. Mark the right angle and the given elevation/depression/bearing.
3. Assign variables to unknown legs; write one tan/sin/cos equation per triangle.
4. For two positions, introduce the shared height and eliminate the unknown base segments.
5. Keep answers in surd form (e.g. 20√3) unless decimals are requested.
6. Sanity-check: larger elevation ⇒ closer or taller; bearing components should not exceed the path length.

Linking Back to Matrices (Optional Insight)

2D rotations by angle θ use the matrix [[cos θ, −sin θ], [sin θ, cos θ]]—determinant 1, so orientation-preserving and invertible. You will not usually multiply rotation matrices on the CAE initial maths paper, but recognising that trig and matrices meet in transformations reinforces why both topics sit in the same engineering maths chapter cluster.

Master elevation/depression sketches and bearing components and you convert lengthy English into two-line trig—exactly what a timed FSc-style engineering entry paper rewards.