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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 |
Andhra Pradesh Board Class 10th Maths Chapter 11 Trigonometry Ex 11.1 Textbooks Solutions PDF
Andhra Pradesh State Board STD 10th Maths Chapter 11 Trigonometry Ex 11.1 Books Solutions with Answers are prepared and published by the Andhra Pradesh Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of AP Board Class 10th Maths Chapter 11 Trigonometry Ex 11.1 Books Answers Solutions, then you are in the right place. Here is a complete hub of Andhra Pradesh State Board Class 10th Maths Chapter 11 Trigonometry Ex 11.1 solutions that are available here for free PDF downloads to help students for their adequate preparation. You can find all the subjects of Andhra Pradesh Board STD 10th Maths Chapter 11 Trigonometry Ex 11.1 Textbooks. These Andhra Pradesh State Board Class 10th Maths Chapter 11 Trigonometry Ex 11.1 Textbooks Solutions English PDF will be helpful for effective education, and a maximum number of questions in exams are chosen from Andhra Pradesh Board.Andhra Pradesh State Board Class 10th Maths Chapter 11 Trigonometry Ex 11.1 Books Solutions
Board | AP Board |
Materials | Textbook Solutions/Guide |
Format | DOC/PDF |
Class | 10th |
Subject | Maths |
Chapters | Maths Chapter 11 Trigonometry Ex 11.1 |
Provider | Hsslive |
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AP Board Class 10th Maths Chapter 11 Trigonometry Ex 11.1 Textbooks Solutions with Answer PDF Download
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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
Answer:
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
Answer:
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
AP Board Textbook Solutions PDF for Class 10th Maths
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