AP Board Class 10 Maths Chapter 11 Trigonometry Ex 11.2 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 11 Trigonometry Ex 11.2 Book Answers ~ HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board

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AP Board Class 10 Maths Chapter 11 Trigonometry Ex 11.2 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 11 Trigonometry Ex 11.2 Book Answers

AP Board Class 10th Maths Chapter 11 Trigonometry Ex 11.2 Textbooks Solutions and answers for students are now available in pdf format. Andhra Pradesh Board Class 10th Maths Chapter 11 Trigonometry Ex 11.2 Book answers and solutions are one of the most important study materials for any student. The Andhra Pradesh State Board Class 10th Maths Chapter 11 Trigonometry Ex 11.2 books are published by the Andhra Pradesh Board Publishers. These Andhra Pradesh Board Class 10th Maths Chapter 11 Trigonometry Ex 11.2 textbooks are prepared by a group of expert faculty members. Students can download these AP Board STD 10th Maths Chapter 11 Trigonometry Ex 11.2 book solutions pdf online from this page.

Andhra Pradesh Board Class 10th Maths Chapter 11 Trigonometry Ex 11.2 Textbooks Solutions PDF

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Andhra Pradesh State Board Class 10th Maths Chapter 11 Trigonometry Ex 11.2 Books Solutions

Board	AP Board
Materials	Textbook Solutions/Guide
Format	DOC/PDF
Class	10th
Subject	Maths
Chapters	Maths Chapter 11 Trigonometry Ex 11.2
Provider	Hsslive

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AP Board Class 10th Maths Chapter 11 Trigonometry Ex 11.2 Textbooks Solutions with Answer PDF Download

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10th Class Maths 11th Lesson Trigonometry Ex 11.2 Textbook Questions and Answers

Question 1.
Evaluate the following.
i) sin 45° + cos 45°
Answer:
sin 45° + cos 45°
= 12√ + 12√
= 1+12√
= 22√
= 2√×2√2√
= √2

ii)

Answer:

iii)

Answer:

iv) 2 tan² 45° + cos² 30° – sin² 60°
Answer:
2 tan² 45° + cos² 30° – sin² 60°
= 2(1)² + (3√2)2 – (3√2)2
= 21 + 34 – 34
= 8+3−34
= 84
= 2

v)

Answer:

Question 2.
Choose the right option and justify your choice.
i) 2tan30∘1+tan245∘
a) sin 60°
b) cos 60°
c) tan 30°
d) sin 30°
Answer:

ii) 1−tan245∘1+tan245∘
a) tan 90°
b) 1
c) sin 45°
d) 0
Answer:
1−tan245∘1+tan245∘ = 1−(1)21+(1)2
= 01+1 = 02 = 0

iii) 2tan30∘1−tan230∘
a) cos 60°
b) sin 60°
c) tan 60°
d) sin 30°
Answer:

Question 3.
Evaluate sin 60° cos 30° + sin 30° cos 60°. What is the value of sin (60° + 30°). What can you conclude?
Answer:
Take sin 60°.cos 30° + sin 30°.cos 60°
= 3√2 . 3√2 + 12 . 12
= (3√)24 + 14
= 34 + 14
= 3+14
= 44 = 1 …… (1)
Now take sin (60° + 30°)
= sin 90° = 1 …….. (2)
From equations (1) and (2), I conclude that
sin (60°+30°) = sin 60° . cos 30° + sin 30° . cos 60°.
i.e., sin (A + B) = sin A . cos B + cos A . sin B

Question 4.
Is it right to say cos (60° + 30°) = cos 60° cos 30° – sin 60° sin 30° ?
Answer:
L.H.S. = cos (60° + 30°)
cos 90° = 0
R.H.S. = cos 60° . cos 30° – sin 60° . sin 30°.
= 12 . 3√2 – 3√2 . 12
= 3√4 – 3√4 = 0
∴ L.H.S = R.H.S
Yes, it is right to say
cos (60°+30°) = cos 60° . cos 30° – sin 60° . sin 30°.
i.e., cos (A + B) = cos A . cos B – sin A . sin B

Question 5.
In right angle triangle △PQR, right angle is at Q and PQ = 6 cms, ∠RPQ = 60°. Determine the lengths of QR and PR.
Answer:
Given that △PQR is a right angled triangle, right angle is at Q and PQ = 6 cm, ∠RPQ = 60°.

tan 60° = Opposite side to ∠𝑃 Adjacent side to ∠𝑃
√3 = 𝑅𝑄6
which gives RQ = 6√3 cm ……. (1)
To find the length of the side RQ, we consider

∴ The length of QR is 6√3 and RP is 12 cm.

Question 6.
In △XYZ, right angle is at Y, YZ = x, and XY = 2x then determine ∠YXZ and ∠YZX.
Answer:
Note: In the problem take
YX = x, and XZ = 2x.

Given that △XYZ is a right angled triangle and right angle at Y, and YX = x and XZ = 2x.
By Pythagoras theorem
XZ² = XY² + YZ²
(2x)² = (x)² + YZ²
4x² = x² + YZ²
YZ² = 4x² – x² = 3x²
YZ = 3𝑥2‾‾‾‾√ = √3x
Now, from the △XYZ
tan X = 𝑋𝑍𝑋𝑌 = 3√𝑥𝑥
tan X = √3 = tan 60°
∴ Angle YXZ is 60°.
tan Z = 𝑋𝑌𝑌𝑍 = 𝑥3√𝑥
tan Z = 13√ = tan 30°
∴ Angle YZX is 30°.
Hence ∠YXZ and ∠YZX are 60° and 30°.

Question 7.
Is it right to say that
sin (A + B) = sin A + sin B? Justify your answer.
Answer:
Let A = 30° and B = 60°
L.H.S = sin (A + B)
= sin (30° + 60°) = sin 90° = 1
R.H.S = sin 30° + sin 60°
= 12 + 3√2
= 3√+12
Hence L.H.S ≠ R.H.S
So, it is not right to say that sin (A + B) = sin A + sin B

AP Board Textbook Solutions PDF for Class 10th Maths

AP Board Class 10 Maths Chapter 1 Real Numbers Textbook Solutions PDF

AP Board Class 10 Maths Chapter 2 Sets Textbook Solutions PDF

AP Board Class 10 Maths Chapter 3 Polynomials Textbook Solutions PDF

AP Board Class 10 Maths Chapter 4 Pair of Linear Equations in Two Variables Textbook Solutions PDF

AP Board Class 10 Maths Chapter 5 Quadratic Equations Textbook Solutions PDF

AP Board Class 10 Maths Chapter 6 Progressions Textbook Solutions PDF

AP Board Class 10 Maths Chapter 7 Coordinate Geometry Textbook Solutions PDF

AP Board Class 10 Maths Chapter 8 Similar Triangles Textbook Solutions PDF

AP Board Class 10 Maths Chapter 9 Tangents and Secants to a Circle Textbook Solutions PDF

AP Board Class 10 Maths Chapter 10 Mensuration Textbook Solutions PDF

AP Board Class 10 Maths Chapter 11 Trigonometry Textbook Solutions PDF

AP Board Class 10 Maths Chapter 12 Applications of Trigonometry Textbook Solutions PDF

AP Board Class 10 Maths Chapter 13 Probability Textbook Solutions PDF

AP Board Class 10 Maths Chapter 14 Statistics Textbook Solutions PDF

Andhra Pradesh Board Class 10th Maths Chapter 11 Trigonometry Ex 11.2 Textbooks for Exam Preparations

Andhra Pradesh Board Class 10th Maths Chapter 11 Trigonometry Ex 11.2 Textbook Solutions can be of great help in your Andhra Pradesh Board Class 10th Maths Chapter 11 Trigonometry Ex 11.2 exam preparation. The AP Board STD 10th Maths Chapter 11 Trigonometry Ex 11.2 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 10th Maths Chapter 11 Trigonometry Ex 11.2 Books State Board syllabus with maximum efficiency.

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Thursday, June 9, 2022