Mathematics (Core)Exam code: 0580

Trigonometry

Pythagoras and Trigonometry

PractiseFlashcards

Sine, Cosine and Tangent

Labelling the triangle

  • The three ratios connect an angle to two of the sides, and which sides they are depends on the angle chosen
  • Pick the angle in question, call it θ, then label the sides from its point of view
    • H, the hypotenuse, opposite the right angle and always the longest
    • O, the side opposite θ
    • A, the side next to θ
  • H never moves, but O and A swap over if a different angle is chosen, so relabel whenever the angle changes
A right-angled triangle labelled from the point of view of the angle theta: the hypotenuse is the longest side facing the right angle, the opposite is the side facing theta, and the adjacent is the side next to it, with the three ratios written underneath as sin equals opposite over hypotenuse, cos equals adjacent over hypotenuse and tan equals opposite over adjacent
Source: SOHCAHTOA by Save My Exams
  • SOHCAHTOA records the three ratios
    • sin θ = O ÷ H
    • cos θ = A ÷ H
    • tan θ = O ÷ A
  • Like Pythagoras, these work only in right-angled triangles

Finding a side

  • Label the triangle, then note which letter belongs to the side you have and which to the side you are after
  • Those two letters pick the ratio: A and H means cosine, O and A means tangent, and O and H means sine
A right-angled triangle with a 23 cm hypotenuse and an 8 cm side, and the unknown angle y at the top: the 23 cm side is marked H and the 8 cm side is marked A from that angle's point of view, so the ratio needed is cosine
Source: SOHCAHTOA by Save My Exams
  • Substitute, then rearrange to make the unknown the subject
A right-angled triangle with the 43 degree angle theta at the bottom left, the 9 cm side beside it marked A and the unknown x cm side facing it marked O, so the ratio linking them is tangent
Source: SOHCAHTOA by Save My Exams
  • Round to three significant figures unless the question says otherwise

Worked example

Finding a side

Find the length x in the triangle below, correct to 3 significant figures.

A right-angled triangle with the right angle at the top left, a vertical side of 9 cm down the left, an angle of 43 degrees at the bottom vertex between that vertical side and the hypotenuse, and the horizontal top side labelled x cm
Source: Trigonometry to find lengths by Save My Exams

Solution:

  • Label the sides from the 43° angle, not from the right angle
  • The 9 cm side runs from the 43° angle to the right angle, so it is the adjacent
  • x is across the triangle from the 43° angle, so it is the opposite
  • Opposite and adjacent together means tangent
  • tan 43° = x ÷ 9
  • Multiply both sides by 9: x = 9 × tan 43°
  • x = 8.3926… = 8.39 cm to 3 significant figures
  • Write the substituted line, x = 9 tan 43°, before evaluating: that is the method mark
  • The hypotenuse is not involved at all here, so it needs no label

Exam tip

Write the first side down

When a side has to be found before the trigonometry, from a perimeter or from another triangle, write it on its own line and mark it on the diagram: that value earns marks by itself. The ratio is then marked on your value, so a line such as tan x = 9 ÷ w, with your value of w, still scores even if w was wrong. So always finish the question.

Worked example

Finding a side first

The diagram shows two right-angled triangles, PQS and RQS. P, Q and R lie on a straight line, and QS is perpendicular to PR. PS = 10 cm, PQ = 6 cm and angle QSR = 50°. Find QR, correct to 3 significant figures.

Two right-angled triangles PQS and RQS sharing the vertical side QS, with P, Q and R on a horizontal line and right angles at Q on both sides. PS is labelled 10 cm, PQ is labelled 6 cm and angle QSR is marked 50°. The length of QS and the length of QR are not given. NOT TO SCALE.

Solution:

  • Triangle RQS has no known length yet, so find QS first, from triangle PQS
  • In triangle PQS the right angle is at Q, so PS = 10 cm is the hypotenuse and QS is a shorter side: subtract
  • QS² = 10² − 6² = 100 − 36 = 64
  • QS = √64 = 8 cm, so write 8 cm on QS in the diagram
  • In triangle RQS, label the sides from the 50° angle at S: QR is opposite and QS is adjacent, so use tangent
  • tan 50° = QR ÷ 8
  • QR = 8 × tan 50° = 9.5340… = 9.53 cm to 3 significant figures

Finding an angle

  • Label the triangle and identify which two sides are known
  • Write the ratio, which leaves the angle as the unknown
  • Apply the inverse function, written sin⁻¹, cos⁻¹ or tan⁻¹, usually reached with the shift key
  • The syllabus states that angle answers should be given correct to one decimal place, so use that unless told otherwise
  • Check the calculator is in degrees before starting, since a wrong mode makes every answer wrong

Worked example

Finding an angle

A right-angled triangle has hypotenuse 13 cm and the side opposite angle θ is 5 cm. Find θ, correct to 1 decimal place.

A right-angled triangle with the right angle at the bottom left. The vertical side is labelled 5 cm and the sloping side is labelled 13 cm. The angle at the bottom right vertex, between the horizontal side and the sloping side, is marked with the Greek letter theta. The horizontal side is unlabelled.

Solution:

  • The known sides are O and H, which is SOH, so use sine
  • sin θ = 5 ÷ 13
  • Apply the inverse: θ = sin⁻¹(5 ÷ 13)
  • θ = 22.61986…
  • So θ = 22.6° to 1 decimal place
  • Check: the third side is 12 by Pythagoras, and the smallest angle faces the shortest side

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