Maths · Trigonometry · chapter 8 of 14
Class 10 Maths Chapter 8: Introduction to Trigonometry case study question with answers
One case-based question of the kind the board asks in Section E: a passage and a figure, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.
Case study · 4 marks
A grain godown in Ludhiana uses a sloping conveyor belt to carry sacks from the ground on to a loading platform. The belt AC starts at a point A on the ground and ends at the edge C of the platform. B is the foot of the platform wall directly below C, so that ∠ABC = 90°. The horizontal distance AB is 4.0 m and the height BC of the platform is 0.9 m, as shown in the figure. The manager wants the length of the belt and the trigonometric ratios of ∠A, the angle the belt makes with the ground.
Part 1 · 1 mark
The length of the belt AC is
Show answer
Answer: (c) 4.1 m
AC = √(4.0² + 0.9²) = √(16 + 0.81) = √16.81 = 4.1 m. (4.9 m adds the two sides; 3.1 m subtracts them.)
Part 2 · 1 mark
The value of tan A is
Show answer
Answer: (b) 940
tan A = BC/AB = 0.9/4.0 = 9/40.
Part 3 · 2 marks
Find sin A and cos A, and verify that sin²A + cos²A = 1.
Answer and marking scheme
sin A = BC/AC = 0.9/4.1 = 9/41 and cos A = AB/AC = 4.0/4.1 = 40/41. sin²A + cos²A = 81/1681 + 1600/1681 = 1681/1681 = 1.
- sin A = 9/41 and cos A = 40/41 (1)
- sum of squares = 1681/1681 = 1 (1)
Or
Find sec A and tan A, and verify that 1 + tan²A = sec²A.
Answer and marking scheme
sec A = AC/AB = 4.1/4.0 = 41/40 and tan A = 9/40. 1 + tan²A = 1 + 81/1600 = 1681/1600 = (41/40)² = sec²A.
- sec A = 41/40 and tan A = 9/40 (1)
- both sides equal 1681/1600 (1)
← Chapter 7: Coordinate GeometryChapter 9: Some Applications of Trigonometry →