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AP Board Class 10 Maths Chapter 11 Trigonometry Ex 11.1 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 11 Trigonometry Ex 11.1 Book Answers

 AP Board Class 10 Maths Chapter 11 Trigonometry Ex 11.1 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 11 Trigonometry Ex 11.1 Book Answers

Andhra Pradesh State Board Class 10th Maths Chapter 11 Trigonometry Ex 11.1 Books Solutions

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Find below the list of all AP Board Class 10th Maths Chapter 11 Trigonometry Ex 11.1 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:

10th Class Maths 11th Lesson Trigonometry Ex 11.1 Textbook Questions and Answers

Question 1.
In right angle triangle ABC, 8 cm, 15 cm and 17 cm are the lengths of AB, BC and CA respectively. Then, find out sin A, cos A and tan A.
Given that
△ABC is a right angle triangle and Lengths of AB, BC and CA are 8 cm, 15 cm and 17 cm respectively.
Among the given lengths CA is longest.
Hence CA is the hypotenuse in △ABC and its opposite vertex having right angle.
i.e., ∠B = 90°.
With reference to ∠A, we have opposite side = BC = 15 cm
adjacent side = AB = 8 cm
and hypotenuse = AC = 17

sin A = Opposite side of ∠A Hypotenuse = 𝐵𝐶𝐴𝐶 = 1517
cos A = Adjacent side of ∠A Hypotenuse = 𝐴𝐵𝐴𝐶 = 817
tan A = Opposite side of ∠A Adjacent side of ∠A = 𝐵𝐶𝐴𝐵 = 158
∴ sin A = 1517;
cos A = 817
tan A = 158

Question 2.
The sides of a right angle triangle PQR are PQ = 7 cm, QR = 25 cm and ∠P = 90° respectively. Then find, tan Q – tan R.

Given that △PQR is a right angled triangle and PQ = 7 cm, QR = 25 cm.
By Pythagoras theorem QR2 = PQ2 + PR2
(25)2 = (7)2 + PR2
PR2 = (25)2 – (7)2 = 625 – 49 = 576
PR = √576 = 24 cm
tan Q = 𝑃𝑅𝑃𝑄 = 247;
tan R = 𝑃𝑄𝑃𝑅 = 724
∴ tan Q – tan R = 247 – 724
= (24)2−(7)2168
= 576−49168
= 527168

Question 3.
In a right angle triangle ABC with right angle at B, in which a = 24 units, b = 25 units and ∠BAC = θ. Then, find cos θ and tan θ.
Given that ABC is a right angle triangle with right angle at B, and BC = a = 24 units, CA = b = 25 units and ∠BAC = θ.

By Pythagoras theorem
AC2 = AB2 + BC2
(25)2 = AB2 + (24)2
AB2 = 252 – 242 = 625 – 576
AB2 = 49
AB = √49 = 1
With reference to ∠BAC = θ, we have
Opposite side to θ = BC = 24 units.
Adjacent side to θ = AB = 7 units.
Hypotenuse = AC = 25 units.
Now
cos θ = Adjacent side of 𝜃 Hypotenuse = 𝐴𝐵𝐴𝐶 = 725
tan θ = Opposite side of 𝜃 Adjacent side of 𝜃 = 𝐵𝐶𝐴𝐵 = 247
Hence cos θ = 725 and tan θ = 247

Question 4.
If cos A = 1213, then find sin A and tan A.
From the identity
sin2 A + cos2 A = 1
⇒ sin2 A = 1 – cos2 A
= 1 – (1213)2
= 1 – 144169
= 169−144169
= 25169
∴ sin A = 25169‾‾‾‾√ = 513

∴ sin A = 513; tan A = 512

Question 5.
If 3 tan A = 4, then find sin A and cos A.
Given 3 tan A = 4
⇒ tan A = 43
From the identify sec2 A – tan2 A = 1
⇒ 1 + tan2 A = sec2 A

If cos A = 35 then from
sin2 A + cos2 A = 1
We can write sin2A = 1 – cos2A
= 1 – (35)2
= 1 – 925
⇒ sin2 A = 1625
⇒ sin A = 45
∴ sin A = 45; cos A = 35

Question 6.
In △ABC and △XYZ, if ∠A and ∠X are acute angles such that cos A = cos X then show that ∠A = ∠X.

In the given triangle, cos A = cos X
⇒ 𝐴𝐶𝐴𝑋 = 𝑋𝐶𝐴𝑋
⇒ AC = XC
⇒ ∠A = ∠X (∵ Angles opposite to equal sides are also equal)

Question 7.
Given cot θ = 78, then evaluate

cot2 θ = (cot θ)2
= (78)2 = 4964 …… (1)

= sec θ + tan θ
So cot θ = 78
⇒ tan θ = 87
⇒ tan2 θ = (87)2 = 6449
From sec2 θ – tan2 θ = 1
⇒ 1 + tan2 θ = sec2 θ

Question 8.
In a right angle triangle ABC, right angle is at B, if tan A = √3, then find the value of
i) sin A cos C + cos A sin C
ii) cos A cos C – sin A sin C
Given, tan A = 3√1
Hence Opposite side Adjacent side =3√1
Let opposite side = √3k and adjacent side = 1k

In right angled △ABC,
AC2 = AB2 + BC2
(By Pythagoras theorem)
⇒ AC2 = (1k)2 + (√3k)2
⇒ AC2 = 1k2 + 3k2
⇒ AC2 = 4k2
∴ AC = 4𝑘2‾‾‾‾√ = 2k

Andhra Pradesh Board Class 10th Maths Chapter 11 Trigonometry Ex 11.1 Textbooks for Exam Preparations

Andhra Pradesh Board Class 10th Maths Chapter 11 Trigonometry Ex 11.1 Textbook Solutions can be of great help in your Andhra Pradesh Board Class 10th Maths Chapter 11 Trigonometry Ex 11.1 exam preparation. The AP Board STD 10th Maths Chapter 11 Trigonometry Ex 11.1 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 10th Maths Chapter 11 Trigonometry Ex 11.1 Books State Board syllabus with maximum efficiency.

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