Bearings and triangle problems

Sketch the situation, label the known information and mark the length or angle you need. In a compound shape, a useful diagonal may turn one difficult problem into two ordinary triangles.

• Right triangle: try Pythagoras or a basic trig ratio.
• SAS or SSS: cosine rule.
• A complete opposite pair: consider the sine rule.
• Two sides and included angle for area: ½ab sin C.

Do not repeatedly round intermediate lengths or angles. Store them and round the final quantity only. Write down any assumptions, such as level ground, straight paths or a vertical tower.

In the same walk, the final position relative to A has east component 4 sin 120° = 2√3 and north component 6 + 4 cos 120° = 4.

tan θ = east / north = √3/2
θ ≈ 40.8934°

C is northeast of A, so its bearing from A is 040.9° (or 041° to the nearest degree). The bearing of A from C is 220.9° (221°). If a point is in another quadrant, use the north/east signs and the sketch; an inverse tangent alone does not choose the quadrant.

A vertical tower stands on level ground. Two observation points are 20 m apart on a straight line on the same side of its base. The nearer elevation is 45°; the farther is 30°. Let x be the nearer horizontal distance and h the height above the observation level.

h/x = tan 45° = 1 ⇒ x = h
h/(x + 20) = tan 30° = 1/√3
√3h = h + 20
h = 10(√3 + 1) ≈ 27.3 m

This assumes equal observation heights, a vertical tower and horizontal ground along the sighting line. If the measuring instrument is above ground, add its height to obtain the tower’s total height.

In a convex quadrilateral ABCD, AB = 5, AD = 8 and angle BAD = 60°. Also BC = 5 and CD = 6. Draw the internal diagonal BD.

BD² = 5² + 8² − 2(5)(8)cos 60° = 49
BD = 7

Now triangle BCD has three known sides. Its angle at C and area follow:

cos C = (5² + 6² − 7²)/(2 × 5 × 6) = 1/5
sin C = √(1 − 1/25) = 2√6/5
Area BCD = 6√6

Triangle ABD has area 10√3, so the whole quadrilateral has area 10√3 + 6√6. To find the other diagonal AC, calculate and add the two angles at B:

angle ABD = cos⁻¹(1/7)
angle DBC = cos⁻¹(19/35)
AC² = 5² + 5² − 2(5)(5)cos(angle ABC)

Use angle ABC = angle ABD + angle DBC without rounding the intermediate angles. This is a chain of ordinary triangle calculations: one diagonal supplies the information needed for the next.

B is 10 km due north of A. A beacon C lies on a bearing of 030° from A and is 7 km from B. Write AC = t. Angle BAC is 30°, so:

7² = 10² + t² − 20t cos 30°
t² − 10√3t + 51 = 0
t = 5√3 ± 2√6

Both values are positive, so there are two possible beacon positions on the same ray. They are 4√6 ≈ 9.80 km apart. The geometry is the same circle–ray ambiguity as in the sine-rule lesson.

A triangle has one side 5, an adjacent side x and included angle 60°. Its perimeter is 12, so the opposite side is 7 − x.

(7 − x)² = 25 + x² − 5x
49 − 14x = 25 − 5x
x = 8/3, opposite side = 13/3
K = ½ × 5 × (8/3) × sin 60° = 10√3/3

Check that both lengths are positive and the strict triangle inequalities hold. With coordinates, calculate side lengths by the distance formula first, or use a horizontal/vertical base and perpendicular height directly.