Applications of Trigonometry

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Applications of 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

A tower stands on top of a cliff. From a point PP on level ground 9595 m from the base of the cliff, the angle of elevation of the top of the cliff is 26°26° and the angle of elevation of the top of the tower is 44°44°. Find the height of the tower, correct to 1 decimal place.
Show the worked solution
  • Let HH m be the height of the cliff and tt m be the height of the tower.
  • Using the angle of elevation to the top of the cliff:
  • tan⁡26°=H95angle of elevation\tan 26° = \dfrac{H}{95} \quad \scriptsize\textit{angle of elevation}
  • H=95×tan⁡26°H = 95 \times \tan 26°
  • H≈95×0.4877H \approx 95 \times 0.4877
  • H=46.33H = 46.33 m (to 2 d.p.)
  • Using the angle of elevation to the top of the tower:
  • tan⁡44°=H+t95angle of elevation\tan 44° = \dfrac{H + t}{95} \quad \scriptsize\textit{angle of elevation}

Answer: 45.445.4

Example 2

A boat sailed 3030 km from a point PP to an island QQ on a bearing of 167°167°. It then sailed another 3030 km on a bearing of 067°067° to a lighthouse RR. Find the distance PRPR, correct to 1 decimal place.
30 km30 km?NNPQRNot drawn to scale
Show the worked solution
  • Draw north lines at both PP and QQ (parallel lines).
  • ∠x=180°−167°\angle x = 180° - 167° (interior ∠\angles, north lines parallel)
  • ∠x=13°\angle x = 13°
  • ∠PQR=13°+67°\angle PQR = 13° + 67°
  • ∠PQR=80°\angle PQR = 80°
  • PR2=PQ2+QR2−2×PQ×QR×cos⁡(∠PQR)cosine rulePR^2 = PQ^2 + QR^2 - 2 \times PQ \times QR \times \cos(\angle PQR) \quad \scriptsize\textit{cosine rule}
  • PR2=302+302−2(30)(30)cos⁡80°PR^2 = 30^2 + 30^2 - 2(30)(30)\cos 80°
  • PR2=900+900−1800cos⁡80°PR^2 = 900 + 900 - 1800\cos 80°

Answer: 38.638.6

Example 3

A cuboid has length 44 cm, width 33 cm and height 99 cm. Find the angle between the diagonal of the base and the space diagonal, correct to 1 decimal place.
?4 cm3 cm9 cmNot drawn to scale
Show the worked solution
  • Step 1: Find the diagonal of the base.
  • d12=42+32d_1^2 = 4^2 + 3^2
  • =25= 25
  • d1=5d_1 = 5 cm base diagonal\quad \scriptsize\textit{base diagonal}
  • Step 2: Find the space diagonal.
  • d2=d12+92d^2 = d_1^2 + 9^2
  • =25+81= 25 + 81
  • =106= 106

Answer: 60.960.9

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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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