This question is inspired from  Question 24 CBSE Class 10 Sample Paper for 2020 Boards

## 'Skysails' is that genre of engineering science that  uses extensive utilization of wind energy to move  a vessel in the sea water. The sky sails technology  allows the towing kite to gain a height of anything  between 100 m to 300 m. The sailing kite is made  in such a way that it can be raised to its proper  elevation and then brought back with the help of  a telescopic mast that enables the kite to be raised  properly and effectively.  Based on the following figure related to sky sailing  answer the questions: ## (c) 60°                             (d) None of these  ## (C)  1                         (D) 1/3 ## (c) 100 m                                 (d) 200 m ## (C) 2                                     (D) 4  ## (c) 3                                                     (d) 4  1. Chapter 8 Class 10 Introduction to Trignometry (Term 1)
2. Serial order wise
3. Case Based Questions (MCQ)

Transcript

Question 'Skysails' is that genre of engineering science that uses extensive utilization of wind energy to move a vessel in the sea water. The sky sails technology allows the towing kite to gain a height of anything between 100 m to 300 m. The sailing kite is made in such a way that it can be raised to its proper elevation and then brought back with the help of a telescopic mast that enables the kite to be raised properly and effectively. Based on the following figure related to sky sailing answer the questions: Question 1 (i) In the given figure, if tan 𝜃 = cot (30° + 𝜃), where θ and 30° + 𝜃 are acute angles, then the value of 𝜃 is: (a) 45° (b) 30° (c) 60° (d) None of these Given, tan θ = cot(30° + 𝜃) tan θ = tan[90° – (30° + 𝜽)] tan θ = tan(90° – 30° – 𝜃) tan θ = tan(60° – 𝜃) Comparing angles θ = 60° – 𝜽 θ + θ = 60° 2θ = 60° θ = (60°)/2 θ = 30° So, the correct answer is (B) 2θ = 60° θ = (60°)/2 θ = 30° So, the correct answer is (B) 2θ = 60° θ = (60°)/2 θ = 30° So, the correct answer is (B) Question 2 The value of tan 30°. cot 60° is: (A) √3 (b) 1/√3 (C) 1 (D) 1/3 tan 30° × cot 60° = 1/√3 × 1/√3 = 𝟏/𝟑 So, the correct answer is (D) Question 3 What should be the length of the rope of the kite sail in order to pull the ship at the angle θ and be at a vertical height of 200 m? (a) 400 m (b) 300 m (c) 100 m (d) 200 m In ∆ABC, given θ = 30°, AB = 200 m Now, sin 30° = 𝐴𝐵/𝐴𝐶 1/2 = 200/𝐴𝐶 AC = 400 m So, the correct answer is (A) Question 4 If cos A = 1/2 , then the value of 9 cot2 A − 1 is : (A) 1 (B) 3 (C) 2 (D) 4 Given, cos A = 1/2 cos A = cos 60° ∴ A = 60° Now, 9 cot2 A − 1 = 9 × cot2 60° − 1 = 9 × (1/√3)^2 − 1 Given, cos A = 1/2 cos A = cos 60° ∴ A = 60° Now, 9 cot2 A − 1 = 9 × cot2 60° − 1 = 9 × (1/√3)^2 − 1 = 9 × 1/3 − 1 = 3 − 1 = 2 So, the correct answer is (C) Question 5 In the given figure, the value of (sin C + cos A) is: (a) 1 (b) 2 (c) 3 (d) 4 In figure ∠ C = θ = 30° And, ∠ B = 90° Now, Sum of angles in Δ ABC = 180° ∠ A + ∠ B + ∠ C = 180° ∠ A + 90° + 30° = 180° ∠ A + 120° = 180° ∠ A = 180° − 90° − 30° ∠ A = 180° − 120° ∠ A = 60° Thus, sin C + cos A = sin 30° + cos 60° = 1/2 + 1/2 = 1 So, the correct answer is (D) 