Class 10 Maths NCERT Exemplar Solution: Introduction to Trigonometry & its Applications (Part-IIA)
Here you will get the class 10 Mathematics chapter 8, Introduction to Trigonometry and its Applications: NCERT Exemplar Problems and Solutions (Part-IIA). This part brings you the solutions Q. No. 1-7 of exercise 8.2 of NCERT Exemplar for Mathematics chapter 8. All the solutions are designed to give you an apt explanation in an easily understandable format.
Here you get the CBSE Class 10 Mathematics chapter 8, Introduction to Trigonometry and its Applications: NCERT Exemplar Problems and Solutions (Part-IIA). This part of the chapter includes solutions of Question Number 1 to 7 from Exercise 8.2 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Introduction to Trigonometry and its Applications. This exercise comprises only the Very Short Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed solution.
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.
Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Introduction to Trigonometry and its Applications:
Very Short Answer Type Questions (Q. No. 1-7)
Write whether True or False and justify your answer.
Question. 2 The value of the expression (cos2 23° - sin2 67°) is positive.
Question. 3 The value of the expression (sin 80° - cos 80°) is negative.
We know that for domain, 0° £ q £90° the value of sinq is increasing and that of cosq is decreasing .
∴ sin80° - cos80°> 0
So, the value of the given expression is positive.
Question. 5 If cos A + cos2 A = 1, then sin2 A + sin4 A = l
Given, cosA + cos2 A = 1
⟹ cos A = 1 - cos2A
⟹ cos A = sin2 A [∵ sin2 A = 1 − cos2A]
⟹ cos2 A = sin4 A [Squaring both sides]
⟹ 1 - sin2 A = sin4 A [∵ cos2 A = 1 - sin2 A]
⟹ sin2 A + sin4 A = 1
Question. 6 (tanθ + 2) (2 tanθ + 1) = 5 tanθ + sec2θ
LHS = (tanθ + 2)(2tanθ + 1)
=2 tan2θ + 4 tanθ + tanθ + 2
=2 (sec2θ -1) + 5 tanθ + 2 [∵ sec2θ -tan2θ = 1]
=2sec2θ + 5tanθ ≠ RHS
Question. 7 If the length of the shadow of a tower is increasing, then the angle of elevation of the Sun is also increasing.
Let BC be the tower and DB = x be the length of initial shadow and DA = y be the the increase in length of shadow.
Also, let the angle of elevation of the sun changes from θ1 to θ2 with the increase in the length of shadow from x to (x + y).
Here, θ1 is the exterior angle of ΔTSD
Therefore, θ1 > θ2.
Hence, the angle of elevation of the Sun is also increasing with an increase in the length of shadow.