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How do you apply trigonometry to bearings, angles of elevation and depression, and problems in three dimensions?

Applying trigonometry, the sine and cosine rules and the area formula to practical problems involving bearings, angles of elevation and depression, and three-dimensional shapes.

A focused answer to the SQA National 5 Mathematics applications of trigonometry content, covering bearings, angles of elevation and depression, and using the sine rule, cosine rule and area formula in practical and three-dimensional problems.

Generated by Claude Opus 4.810 min answer

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  1. What this dot point is asking
  2. Bearings
  3. Angles of elevation and depression
  4. Choosing the right tool
  5. Three-dimensional problems
  6. Examples in context
  7. Try this

What this dot point is asking

The SQA wants you to apply trigonometry to real problems: bearings, angles of elevation and depression, and shapes in three dimensions. You must read a worded problem, draw and label the right triangle, and choose between basic trigonometry, the sine rule, the cosine rule and the area formula.

Bearings

A bearing fixes a direction as an angle measured clockwise from north, always written with three figures. So due east is 090∘090^\circ, due south is 180∘180^\circ, and north-west is 315∘315^\circ. Bearing problems usually form a triangle from two journey legs, and the angle inside the triangle comes from the difference of the bearings and the parallel north lines.

Angles of elevation and depression

The angle of elevation is the angle measured upward from the horizontal to an object above you; the angle of depression is measured downward to an object below you. These angles are equal when looking between two points (they are alternate angles between parallel horizontals).

Choosing the right tool

Real problems rarely tell you which rule to use, so the decision is part of the skill. If the triangle has a right angle, use sine, cosine or tangent, or Pythagoras. If it does not, use the sine rule when a side and its opposite angle are known, and the cosine rule when two sides and the included angle, or three sides, are known.

Three-dimensional problems

In three dimensions, find a right-angled triangle in a plane within the solid and work in that plane. Often you find one length in a horizontal triangle, then use it as a side of a vertical triangle to find an angle of elevation or a slant length, exactly as with Pythagoras in three dimensions.

Examples in context

Surveyors, sailors and pilots use these methods daily. A coastguard fixes a boat's position from two bearings taken at different points, then solves the triangle with the sine rule. A builder finds the length of a sloping roof valley using a horizontal diagonal and a vertical rise. Navigation between waypoints relies on bearings and the cosine rule to find direct distances.

Try this

Q1. A bearing is sixty-five degrees clockwise from north. Write it correctly. [1 mark]

  • Cue. 065∘065^\circ.

Q2. From 4040 m away, the angle of elevation to a tree top is 30∘30^\circ. Find its height. [2 marks]

  • Cue. 40tan⁑30∘=23.140 \tan 30^\circ = 23.1 m.

Q3. A triangle has sides 55 km and 77 km with a 60∘60^\circ angle between them. Find the third side. [3 marks]

  • Cue. d2=25+49βˆ’70cos⁑60∘=39d^2 = 25 + 49 - 70\cos 60^\circ = 39, so d=6.2d = 6.2 km.

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA National 5 20194 marksA ship sails 3030 km on a bearing of 060∘060^\circ, then 4040 km on a bearing of 150∘150^\circ. Calculate its straight-line distance from the start.
Show worked answer β†’

The two bearings differ by 150βˆ’60=90∘150 - 60 = 90^\circ, and because the second leg turns through the angle between the legs the included angle of the triangle is 90∘90^\circ (1 mark). With a right angle, use Pythagoras: d2=302+402=900+1600=2500d^2 = 30^2 + 40^2 = 900 + 1600 = 2500 (2 marks). So d=2500=50d = \sqrt{2500} = 50 km (1 mark). Markers reward finding the included angle from the bearings, the right-angle relationship, and the distance.

SQA National 5 20223 marksFrom a point 5050 m from the base of a tower, the angle of elevation to the top is 35∘35^\circ. Calculate the height of the tower, to 1 decimal place.
Show worked answer β†’

The height is the opposite side and 5050 m is the adjacent side, so use the tangent ratio: tan⁑35∘=h50\tan 35^\circ = \dfrac{h}{50} (1 mark). Rearrange: h=50tan⁑35∘h = 50 \tan 35^\circ (1 mark). Evaluate: h=50Γ—0.7002=35.0h = 50 \times 0.7002 = 35.0 m to 1 decimal place (1 mark). Markers reward choosing the tangent ratio, rearranging, and the rounded height.

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