National 5 Mathematics: geometry and trigonometry

The shape and trigonometry questions in National 5 Maths mostly come down to picking the right formula and substituting carefully. This set covers Pythagoras, the area of a triangle from two sides and the angle between them, the sine rule, the cosine rule for both a missing side and a missing angle, arc length and sector area, volume scale factors for similar solids, the volume of a cone, and a trigonometric equation with two solutions in the range.

Every measurement is written into the question, so nothing depends on a diagram you cannot see. The wrong options are the errors that actually lose marks: the two short sides added instead of squared, the square root forgotten, the sine-rule ratio turned upside down, the cosine term added rather than subtracted, the diameter used where the radius belongs, arc length given when the area was asked for, and a linear scale factor applied to a volume. In the cosine-rule angle question the three wrong answers are the other two angles of the same triangle and its supplement, so guessing by size does not help.

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  • Find a hypotenuse with Pythagoras and know when the answer still needs a square root
  • Calculate the area of a triangle from two sides and the angle between them
  • Use the sine rule the right way up to find a missing side
  • Use the cosine rule to find a side when two sides and the included angle are known
  • Rearrange the cosine rule to find an angle from three sides
  • Calculate arc length and sector area as fractions of a full circle
  • Apply the cube of the linear scale factor when comparing volumes of similar solids
  • Find the volume of a cone without dropping the one third
  • Solve a trigonometric equation and give every solution in the stated range

A triangle has sides of 5 cm, 6 cm and 7 cm. What is the size of the angle opposite the 7 cm side? — $\cos(A) = \frac{5^2 + 6^2 - 7^2}{2(5)(6)} = \frac{12}{60} = 0.2$, so $A \approx 78.5$ degrees.

Sample question

A right-angled triangle has its two shorter sides measuring 9 cm and 12 cm. What is the length of the hypotenuse?

See the answer

15 cm

Using Pythagoras' theorem, $a^2 + b^2 = c^2$, we have $9^2 + 12^2 = 81 + 144 = 225$. The hypotenuse is $\sqrt{225} = 15$ cm. Adding the sides (21 cm) is a common error, as is forgetting the square root (225 cm).

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