Finding Ratios from a Given Ratio and Quadrant

Finding Ratios from a Given Ratio and Quadrant

Secondary 4
TGM Original Questions

A common exam task gives one ratio together with a quadrant range and asks for other ratios exactly, without a calculator. Build a right-angled triangle in the correct quadrant and read everything off.

Method: Quadrant, Triangle, Pythagoras
  1. Locate the quadrant from the given angle range.
  2. Draw the rotating arm in that quadrant and drop a perpendicular to the \(x\)-axis to form a right-angled triangle.
  3. Fill in two sides from the given ratio, with signs from the quadrant: \(x\) negative on the left, \(y\) negative below; \(r\) is always positive.
  4. Use Pythagoras' theorem to find the third side.
  5. Read off the required ratios and check each sign against the ASTC rule.

Setup for \(\tan\theta=-\dfrac{9}{40}\) with \(90^\circ<\theta<180^\circ\)

\(x=-40,\quad y=9\)

\(r=\sqrt{40^2+9^2}=41\)

\(\sin\theta=\dfrac{9}{41}\)

\(\cos\theta=-\dfrac{40}{41}\)

\(\sec\theta=-\dfrac{41}{40}\)

The second quadrant makes \(x\) negative and \(y\) positive — exactly why \(\tan\theta\) is negative there while \(\sin\theta\) stays positive.

Exam check

Check the sign of every final ratio against the quadrant before moving on — the triangle gives sizes, the quadrant gives signs.

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