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Trigonometric Ratios

Secondary 2 practice

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Trigonometric Ratios practice questions

Secondary 2 topics, working towards the O-Level Mathematics 4052 syllabus. Every question below has a full worked solution.

When a student gets a question wrong in Math Amigo, it does not show the answer straight away. It reads the working, identifies the step that went wrong, and asks a question that points at the correction. The worked solution is there once a genuine attempt has been made.

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What this topic covers

Worked examples

Example 1

In △ABC, ∠ACB=90°\angle ACB = 90°, ∠A=59°\angle A = 59° and AC=15AC = 15 cm. Find the length of ABAB, correct to the nearest whole number.
59°ABC?15 cmNot drawn to scale
Show the worked solution
  • cos⁡59°=ACABadjacent over hypotenuse\cos 59° = \dfrac{AC}{AB} \quad \scriptsize\textit{adjacent over hypotenuse}
  • AB×cos⁡59°=ACmultiply both sides by ABAB \times \cos 59° = AC \quad \scriptsize\textit{multiply both sides by } AB
  • AB=15cos⁡59°divide both sides by cos⁡59°AB = \dfrac{15}{\cos 59°} \quad \scriptsize\textit{divide both sides by } \cos 59°
  • AB=29AB = 29 cm (to the nearest whole number)

Answer: 2929

Example 2

In △ABC, ∠ACB=90°\angle ACB = 90°, AB=74AB = 74 cm and BC=24BC = 24 cm. Find ∠A\angle A, correct to the nearest degree.
?ABC24 cm74 cmNot drawn to scale
Show the worked solution
  • Use Pythagoras’ theorem to find AC:\text{Use Pythagoras' theorem to find } AC:
  • AB2=BC2+AC2AB^2 = BC^2 + AC^2
  • AC2=742−242AC^2 = 74^2 - 24^2
  • AC2=5476−576AC^2 = 5476 - 576
  • AC2=4900AC^2 = 4900
  • AC=70AC = 70 cm
  • tan⁡∠A=BCACopposite over adjacent\tan \angle A = \dfrac{BC}{AC} \quad \scriptsize\textit{opposite over adjacent}
  • tan⁡∠A=2470\tan \angle A = \dfrac{24}{70}

Answer: 1919

Example 3

A tower 88 m tall casts a shadow on level ground. The angle of elevation of the sun is 31°31°. Find the length of the shadow, correct to the nearest metre.
Show the worked solution
  • tan⁡31°=heightshadow lengthopposite over adjacent\tan 31° = \dfrac{\text{height}}{\text{shadow length}} \quad \scriptsize\textit{opposite over adjacent}
  • tan⁡31°=8s\tan 31° = \dfrac{8}{s}
  • s=8tan⁡31°rearranges = \dfrac{8}{\tan 31°} \quad \scriptsize\textit{rearrange}
  • s=13s = 13 m (to the nearest metre)

Answer: 1313

Questions parents ask

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Which syllabus does this follow?

O-Level Mathematics (4052), for Singapore secondary students. Math Amigo covers Secondary 1 to 4, both O-Level Mathematics and Additional Mathematics.

Will it just show my child the answer?

No. On a wrong answer it looks at the working, points to the step that went wrong, and asks a question that helps the student find the correction. The full worked solution is available after a genuine attempt.

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