Here you get the CBSE Class 10 Mathematics chapter 8, Introduction to Trigonometry and its Applications: NCERT Exemplar Problems and Solutions (Part-IIIB). This part is a continuation of the Part-IIIA of Class 10 Maths NCERT Exemplar Problems and Solutions for chapter -8. It includes solutions to Question Number 8 to 15 from Exercise 8.3 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Introduction to Trigonometry and its Applications. This exercise comprises only the Short Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed solution.
Class 10 Maths NCERT Exemplar Solution: Introduction to Trigonometry & its Applications (Part-IIIA)
NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams.
CBSE Class 10 Mathematics Syllabus 2017-2018
Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Introduction to Trigonometry and its Applications:
Exercise 8.3
Short Answer Type Questions (Q. No. 8-15)
Question. 10 A ladder 15m long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, then find the height of the wall.
Solution.
Let AB be the 15 m long ladder making a 60o angle with a h metre high wall AC.
Question. 11 Simplify (1 + tan2 θ) (1 - sin θ) (1 + sin θ).
Solution.
Question. 12 If 2sin2θ - cos2θ = 2, then find the value of θ.
Solution.
Question. 14 An observer 1.5m tall is 20.5m away from a tower 22m high. Determine the angle of elevation of the top of the tower from the eye of the observer.
Solution.
Let AB be the 22 m high tower and CD be the 1.5 m tall observer standing at a distance of BC = 20.5 m away from the tower.
Also, let θ be the angle of elevation of the top of the tower from the eye of the observer.
Question. 15 Show that tan4θ + tan2 θ = sec4 θ - sec2 θ.
Solution.
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