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Further Trigonometry

Secondary 3 practice

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Further Trigonometry practice questions

Secondary 3 topics from 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, AB=12AB = 12 cm, BC=8BC = 8 cm and the area of △ABC is 41.641.6 cm². Find the acute angle ∠ABC\angle ABC. Give your answer correct to 1 decimal place.
?12 cm8 cmABCNot drawn to scale
Show the worked solution
  • Area of △ABC=12×AB×BC×sin⁡(∠ABC)area formula\text{Area of }\triangle ABC = \frac{1}{2} \times AB \times BC \times \sin(\angle ABC) \quad \scriptsize\textit{area formula}
  • 41.6=12×12×8×sin⁡(∠ABC)41.6 = \frac{1}{2} \times 12 \times 8 \times \sin(\angle ABC)
  • 41.6=48sin⁡(∠ABC)simplify41.6 = 48 \sin(\angle ABC) \quad \scriptsize\textit{simplify}
  • sin⁡(∠ABC)=41.648\sin(\angle ABC) = \frac{41.6}{48}
  • ∠ABC=sin⁡−1 ⁣(41.648)\angle ABC = \sin^{-1}\!\left(\frac{41.6}{48}\right)
  • =60.1°= 60.1° (to 1 d.p.)

Answer: 60.160.1

Example 2

In △ABCABC, ∠A=78°\angle A = 78°, AB=11AB = 11 cm and BC=16BC = 16 cm. Find ∠ACB\angle ACB, correct to 1 decimal place.
78°?11 cm16 cmABCNot drawn to scale
Show the worked solution
  • sin⁡∠ACBAB=sin⁡ABCsine rule\dfrac{\sin \angle ACB}{AB} = \dfrac{\sin A}{BC} \quad \scriptsize\textit{sine rule}
  • sin⁡∠ACB11=sin⁡78°16\dfrac{\sin \angle ACB}{11} = \dfrac{\sin 78°}{16}
  • sin⁡∠ACB=11×sin⁡78°16\sin \angle ACB = \dfrac{11 \times \sin 78°}{16}
  • sin⁡∠ACB=0.6725\sin \angle ACB = 0.6725 (to 4 d.p.)
  • ∠ACB=sin⁡−1(0.6725)\angle ACB = \sin^{-1}(0.6725)
  • ∠ACB=42.3°\angle ACB = 42.3° (to 1 d.p.)

Answer: 42.342.3

Example 3

In △ABC\triangle ABC, a=6a = 6 cm, b=16b = 16 cm and c=15c = 15 cm. Find the largest angle of the triangle, correct to 1 decimal place.
?6 cm16 cm15 cmABCNot drawn to scale
Show the worked solution
  • The largest angle is opposite the longest side. Longest side=b=16 cm\text{The largest angle is opposite the longest side. Longest side} = b = 16 \text{ cm}
  • cos⁡B=a2+c2−b22accosine rule rearranged\cos B = \dfrac{a^2 + c^2 - b^2}{2ac} \quad \scriptsize\textit{cosine rule rearranged}
  • cos⁡B=62+152−1622(6)(15)\cos B = \dfrac{6^2 + 15^2 - 16^2}{2(6)(15)}
  • cos⁡B=36+225−256180\cos B = \dfrac{36 + 225 - 256}{180}
  • cos⁡B=136\cos B = \dfrac{1}{36}
  • B=cos⁡−1 ⁣(136)B = \cos^{-1}\!\left(\dfrac{1}{36}\right)
  • B=88.4°B = 88.4° (to 1 d.p.)

Answer: 88.488.4

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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