Class 10 Maths NCERT Exemplar Solution: Introduction to Trigonometry & its Applications (Part-IIA)

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.

CBSE Class 10 Mathematics Syllabus 2017-2018

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:

Exercise 8.2

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.

Solution. False

Question. 3 The value of the expression (sin 80° - cos 80°) is negative.

Solution. False

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

Solution. True

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

Solution. False

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.

Solution. False

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.  

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