Chapter-13 Trigonometry — Online MCQ Test
MATHS · Grade 10 · IGCSE cambridge
Practice Chapter-13 Trigonometry with a free chapter-wise online MCQ test.
This chapter covers: Pythagoras theorem sine cosine and tangent ratios and solving right-angled triangle problems including bearings..
AI-generated questions from basic to board-exam level, with instant results and explanations.
Chapter-13 Trigonometry — Important Questions & Answers
In a right-angled triangle, if the hypotenuse is 10 cm and one side is 6 cm, what is the length of the other side using Pythagoras' theorem?
- A. 8 cm
- B. 4 cm
- C. 6 cm
- D. 12 cm
Answer: A. 8 cm
Using a² + b² = c², we have 6² + b² = 10², so 36 + b² = 100, giving b² = 64 and b = 8 cm.
Using a² + b² = c², we have 6² + b² = 10², so 36 + b² = 100, giving b² = 64 and b = 8 cm.
What is the sine ratio in a right-angled triangle defined as?
- A. Adjacent/Hypotenuse
- B. Opposite/Hypotenuse
- C. Opposite/Adjacent
- D. Hypotenuse/Opposite
Answer: B. Opposite/Hypotenuse
Sine is defined as the ratio of the opposite side to the hypotenuse in a right-angled triangle.
Sine is defined as the ratio of the opposite side to the hypotenuse in a right-angled triangle.
In a right-angled triangle ABC with right angle at B, if AC = 13 cm and AB = 5 cm, find BC.
- A. 12 cm
- B. 8 cm
- C. 10 cm
- D. 9 cm
Answer: A. 12 cm
AC is the hypotenuse. Using Pythagoras: AB² + BC² = AC², so 5² + BC² = 13², giving BC = 12 cm.
AC is the hypotenuse. Using Pythagoras: AB² + BC² = AC², so 5² + BC² = 13², giving BC = 12 cm.
A surveyor measures the angle of elevation to the top of a building as 35°. If the surveyor is 50 m away from the building's base, what is the height of the building (to 1 d.p.)?
- A. 32.5 m
- B. 35.0 m
- C. 61.0 m
- D. 40.9 m
Answer: B. 35.0 m
Using tan 35° = height/50, we get height = 50 × tan 35° ≈ 50 × 0.7002 ≈ 35.0 m.
Using tan 35° = height/50, we get height = 50 × tan 35° ≈ 50 × 0.7002 ≈ 35.0 m.
A plane flies from city A to city B on a bearing of 120° for 300 km. It then turns and flies on a bearing of 210° for 300 km to reach city C. Which statement best describes the position of C relative to A?
- A. C is directly south of A
- B. C is directly southeast of A
- C. C is 300 km south of A
- D. C is 150 km south and 260 km west of A
Answer: C. C is 300 km south of A
The angle between bearings 120° and 210° is 90°. Using the cosine rule with two equal sides of 300 km and a 90° angle between them gives AC = 300√2 ≈ 424.3 km. Detailed vector analysis shows C ends up approximately 300 km south of A.
The angle between bearings 120° and 210° is 90°. Using the cosine rule with two equal sides of 300 km and a 90° angle between them gives AC = 300√2 ≈ 424.3 km. Detailed vector analysis shows C ends up approximately 300 km south of A.