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Trigonometry and Bearings

Mathematics · WAEC and JAMB · SS2 and SS3

Trigonometry carries one or two full Paper 2 questions, usually a bearing problem with a diagram and an elevation problem. The marks are mostly for drawing a correct labelled diagram, so candidates who sketch properly score even when the arithmetic wobbles.

What you need to know

  • In a right-angled triangle, sine = opposite/hypotenuse, cosine = adjacent/hypotenuse and tangent = opposite/adjacent. Label the three sides relative to the angle you are using BEFORE choosing the ratio.
  • The hypotenuse is always the side facing the right angle and never changes, but which side is opposite and which is adjacent depends on the angle being used.
  • Learn the exact values: sin 30 = 1/2, cos 30 = sqrt(3)/2, tan 30 = 1/sqrt(3), sin 45 = cos 45 = 1/sqrt(2), tan 45 = 1, sin 60 = sqrt(3)/2, cos 60 = 1/2, tan 60 = sqrt(3). Questions using these expect exact surd answers, not decimals.
  • The identity sin^2 A + cos^2 A = 1 and the relation tan A = sin A / cos A let you find one ratio from another without ever finding the angle itself.
  • An angle of elevation is measured UP from the horizontal to the line of sight, and an angle of depression is measured DOWN from the horizontal. They are alternate angles, so they are equal for the same pair of points.
  • In elevation problems involving a person, remember to add the observer's eye height at the end. The tangent gives you the height ABOVE eye level, not above the ground.
  • A bearing is always measured clockwise from NORTH, written with three digits: due east is 090, due south is 180 and due west is 270.
  • To find a back bearing, add 180 if the forward bearing is less than 180, and subtract 180 if it is 180 or more. The bearing of A from B is the back bearing of B from A.
  • Draw a north line at EVERY point in a bearing question, not just the first. Most bearing errors come from a missing north arrow at the second station.
  • Where two bearings differ by exactly 90 degrees, the triangle formed is right-angled and you can use Pythagoras and simple ratios instead of the cosine rule.
  • The sine rule, a/sin A = b/sin B = c/sin C, is used when you have two angles and a side, or two sides and a non-included angle.
  • The cosine rule, a^2 = b^2 + c^2 - 2bc cos A, is used when you have two sides and the INCLUDED angle, or all three sides and need an angle.
  • The area of any triangle given two sides and the included angle is (1/2)ab sin C, which avoids having to find a perpendicular height.
  • Always check that your angle of depression answer is less than 90 degrees and that the longest side faces the largest angle; that single check catches most calculator-mode errors.

Key terms

Angle of elevation
The angle between the horizontal and the line of sight to an object above the observer.
Angle of depression
The angle between the horizontal and the line of sight to an object below the observer.
Bearing
The direction of one point from another, measured clockwise from north and written with three digits.
Back bearing
The bearing of the starting point measured from the destination, differing from the forward bearing by 180 degrees.
Sine rule
The relation that in any triangle the ratio of each side to the sine of its opposite angle is constant.
Cosine rule
The relation linking all three sides of a triangle with the cosine of one angle, used when the sine rule cannot start.

Formulae

  • sin A = opposite/hypotenuse; cos A = adjacent/hypotenuse; tan A = opposite/adjacent
  • sin^2 A + cos^2 A = 1
  • tan A = sin A / cos A
  • sine rule: a/sin A = b/sin B = c/sin C
  • cosine rule: a^2 = b^2 + c^2 - 2*b*c*cos A
  • cos A = (b^2 + c^2 - a^2) / (2*b*c)
  • area of triangle = (1/2) * a * b * sin C
  • back bearing = forward bearing + 180 (if less than 180) or forward bearing - 180 (if 180 or more)

Worked examples

A man 1.7 m tall stands 20 m from the foot of a mast on level ground. The angle of elevation of the top of the mast from his eyes is 38 degrees. Find the height of the mast, correct to one decimal place. (tan 38 = 0.7813)

  1. Sketch the situation: the horizontal line from his eyes to the mast is 20 m, and the height above eye level is unknown, call it x.
  2. The side opposite the 38 degree angle is x and the adjacent side is 20, so use tangent: tan 38 = x/20.
  3. So x = 20 x tan 38 = 20 x 0.7813 = 15.626 m.
  4. This is the height ABOVE the man's eye level only.
  5. Add his own height: total = 15.626 + 1.7 = 17.326 m.
  6. To one decimal place, the mast is 17.3 m tall.

In triangle ABC, b = 8 cm, c = 5 cm and angle A = 60 degrees. Find the length of side a.

  1. Angle A lies between sides b and c, so it is the included angle and the cosine rule applies.
  2. Write the rule: a^2 = b^2 + c^2 - 2bc cos A.
  3. Substitute: a^2 = 8^2 + 5^2 - 2(8)(5) cos 60.
  4. Since cos 60 = 0.5, the last term is 80 x 0.5 = 40.
  5. So a^2 = 64 + 25 - 40 = 49.
  6. Therefore a = sqrt(49) = 7 cm.
  7. Check: the side of 7 cm lies between 5 and 8, which is sensible for a 60 degree included angle.

A boat sails 12 km from P to Q on a bearing of 040 degrees, then 15 km from Q to R on a bearing of 130 degrees. Find the distance PR and the bearing of R from P, correct to the nearest whole number.

  1. Draw north lines at both P and Q. The bearing changes from 040 to 130, a difference of 90 degrees, so angle PQR is 90 degrees.
  2. Triangle PQR is right-angled at Q, so use Pythagoras: PR^2 = 12^2 + 15^2 = 144 + 225 = 369.
  3. PR = sqrt(369) = 19.209, which is 19.2 km to one decimal place, or 19 km to the nearest whole number.
  4. For the bearing, find angle QPR: tan(QPR) = opposite/adjacent = 15/12 = 1.25.
  5. So angle QPR = 51.3 degrees (since tan 51.3 is about 1.248).
  6. The bearing of R from P = bearing of Q from P plus angle QPR = 040 + 51.3 = 091.3 degrees.
  7. To the nearest whole number the bearing is 091 degrees.

The mistake to avoid

Candidates give the height above eye level as the full height of the building or mast, forgetting to add the observer's height. In bearings, the other standing error is writing a bearing as 51 degrees instead of 091 degrees, or omitting the leading zero in a two-digit bearing.

In the exam

Draw the diagram large, mark the north line at every station, and label all known distances and angles before you write a formula. State clearly which rule you are using and why, since 'two sides and included angle, so cosine rule' earns credit. Give bearings in three figures with the degree sign, and keep your calculator in degree mode.