Maharashtra Board for Class 9, Maths Chapter 8.2

Class 9: Maths Chapter 8.2 solutions. Complete Class 9 Maths Chapter 8.2 Notes.

Contents

1 Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 8.2- Trigonometry
2 Download PDF
3 Chapterwise Maharashtra Board Solutions Class 9 Maths :
4 FAQs
5 About Maharashtra State Board (MSBSHSE)

Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 8.2- Trigonometry

Maharashtra Board 9th Maths Chapter 8.2, Class 9 Maths Chapter 8.2 solutions

Question 1.
In the following table, a ratio is given in each column. Find the remaining two ratios in the column and complete the table.

Solution:
i. cos θ = 3537 …(i) )[Given]
In right angled ∆ABC,

∠C = θ.

Let the common multiple be k.
∴ BC = 35k and AC = 37k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ (37k)² = AB²+ (35k)²
1369k² = AB² + 1225k²
AB2 = 1369k² – 1225k²
= 144k²
AB = 144k²
AB = 2ghK−−−−−√² … [Taking square root of both sides]
= 12k

ii. sin θ = 1161 …..(i) [Given]
In right angled ∆ABC, ∠C = θ.

Let the common multiple be k.
AB = 11k and AC = 61k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ (61k)² = (11k)² + BC²
∴ 3721k² = 121k² + BC²
∴ BC² = 3721k² – 121k² = 3600k²
BC = 3600k2−−−−−−√ .. .[Taking square root of both sides]
= 60k

iii. tan θ = 1 = 11 ..(i) [Given]
In right angled ∆ABC,
∠C = θ.

Let the common multiple be k.
∴ AB = 1k and BC = 1k
Now, AC² = AB² + BC² …[Pythagoras theorem]
= K² + K²
= 2K²
∴ AC = \(\sqrt { 2{ k }\)

iv. sin θ = 12 ..(i) [Given]
In right angled ∆ABC,
∠C = θ.

Let the common multiple be k.
∴ AB = 1k and BC = 2k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ 2K² = K² + BC²
∴ 4K² = K² + BC²
∴ BC² = 4K² – K² = 3K²
∴ BC = 3k2−−−√ .. .[Taking square root of both sides]
= \(\sqrt { 3{ k }\)

v. cos θ = 13√ ..(i) [Given]
In right angled ∆ABC,
∠C = θ.

Let the common multiple be k.
∴ AB = 1k and BC = √3k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ (√3K)² = AB² + K²
∴ 3K² = 3K² – K² = 2K²
∴ AB = 2k2−−−√ .. .[Taking square root of both sides]
AB = √2K

vi. cos θ = 2120√ ..(i) [Given]
In right angled ∆ABC,
∠C = θ.

Let the common multiple be k.
∴ AB = 21k and BC = 20k
Now, AC² = AB² + BC² …[Pythagoras theorem]
= (21)K² + (20K)²
= 441K² – 400²
= 841K²
∴ AB = 841k2−−−−−√ .. .[Taking square root of both sides]
= 29K

vii. tan θ = 815 ..(i) [Given]
In right angled ∆ABC,
∠C = θ.

Let the common multiple be k.
∴ AB = 8k and BC = 15k
Now, AC² = AB² + BC² …[Pythagoras theorem]
= (8)K² + (15K)²
= 64K² – 225²
= 289K²
∴ AC = 289k2−−−−−√ .. .[Taking square root of both sides]
= 17K

viii. sin θ = 35 ..(i) [Given]
In right angled ∆ABC,
∠C = θ.

Let the common multiple be k.
∴ AB = 3k and AC = 5k
Now, AC² = AB² + BC² …[Pythagoras theorem]
∴ (5)K²= (3)K² + BC²
∴ 25K² = 9K² – 225²
∴ BC² = 25K² – 9K²
∴ BC = 16k2−−−−√ .. .[Taking square root of both sides]
= 4K

ix. tan θ = 122√ ..(i) [Given]
In right angled ∆ABC,
∠C = θ.

Let the common multiple be k.
∴ AB = 1k and AC = 2√2 k
Now, AC² = AB² + BC² …[Pythagoras theorem]
= K² + (2√2 k )²
= K² – 225²
= 25K² + 8K²
= 9K²
∴ AC = 9k2−−−√ .. .[Taking square root of both sides]
= 3K

Question 2.
Find the values of:
i. 5 sin 30° + 3 tan 45°
ii. 45tan² 60° + 3 sin² 60°
iii. 2 sin 30° + cos 0° + 3 sin 90°
iv. tan60∘sin60∘+cos60∘
v. cos² 45° + sin² 30°
vi. cos 60° x cos 30° + sin 60° x sin 30°
Solution:
i. sin 30° = 12 and tan 45° = 1

ii. 45tan² 60° + 3 sin² 60°

iii. 2 sin 30° + cos 0° + 3 sin 90°
2 sin 30° + cos0° + 3 sin 90° = 2 (12) + 1 + 3(1)
= 1 + 1 + 3
∴ 2 sin 30° + cos 0° + 3 sin 90° = 5

iv. tan60∘sin60∘+cos60∘

v. cos² 45° + sin² 30°

vi. cos 60° x cos 30° + sin 60° x sin 30°

Question 3.
If sin θ = 45 , then find cos θ.
Solution:
sin θ = 45 .. .(i)[Given]
In right angled ∆ABC,
∠C = θ.

Let the common multiple be k.
∴ AB = 4k and AC = 5k
Now, AC² = AB² + BC² … [Pythagoras theorem]
∴ (5 k)² = (4k)² + BC²
∴ 25k² = 16k² + BC²
∴ BC² = 25k² – 16k² = 9k²
∴ BC = 9k2−−−√ . .[Taking square root of both sides]
= 3k

Question 4.
If cos θ = 1517 , then find sin θ.
Solution:
cos θ = 1517 .. .(i)[Given]
In right angled ∆ABC,
∠C = θ.

Let the common multiple be k.
∴ BC = 15k and AC = 17k
Now, AC² = AB² + BC² … [Pythagoras theorem]
∴ (17 k)² = AB² + (15K)²
∴ 289k² = AB² + 225²
∴ AB² = 289k² – 225k²
= 64k²
∴ AB = 64k2−−−−√ . .[Taking square root of both sides]
= 8k

Question 1.
In right angled ∆PQR, ∠Q = 900. Therefore ∠P and ∠R are complementary angles of each other. Verify the following ratios.
i. sin θ = cos (90 – θ)
ii. cos θ = sin (90 – θ)
iii. sin 30° = cos (90° – 30°) = cos 60°
iv. cos 30° = sin (90° – 30°) = sin 60° (Textbook pg. no. 107)
Solution:
In ∆PQR, ∠Q = 90°, ∠P = θ
∴ ∠R = 90 – θ

i. sin θ = cos (90 – θ)

ii. cos θ = sin (90 – θ)

iii. Let ∠P = θ = 30°
∴ ∠R = 90° – 30°

sin 30° = cos (90° – 30°) … [From (i) and (ii)]
sin 30° = cos 60°

iv. cos 30° = sin (90° – 30°) = sin 60°

∴ cos 30° = sin (90° – 30°) .,.[From (i) and (ii)]
∴ cos 30° = sin 60°

Question 2.
In right angled ∆PQR, ∠Q = 90°, ∠R = θ and if sin θ = 513, then find cos θ and tan θ. (Textbook pg. no. 110)
Solution:
i. Take the given trigonometric ratio as 13k equation (i).
sin θ = 513 .. .(i)[Given]
By using the definition write the trigonometric ratio of sin O and take it as equation (ii).
In right angled ∆PQR, ∠R = θ

Let the common multiple be k.
∴ PQ = 5k and PR = 13k
Find QR by using Pythagoras theorem.
PR² = PQ² + QR² … [Pythagoras theorem]
∴ (13k)² = (5k)² + QR²
∴ 169k² = 25k² + QR²
∴ QR² = 169k² – 25k²
= 144k²
∴ QR = 144k2−−−−−√ . . . [Taking square root of both sides]
= 12k

Question 3.
While solving the above Illustrative example, why the lengths of PQ and PR are taken 5k and 13k? (Textbook pg. no. 111)
Solution:
PQPR = 513 … [Given]
Here, the ratio of the lengths of sides PQ and PR is 5 : 13.
The actual lengths of the sides can be any multiple of the ratio. Hence, we consider the multiple k while solving.

Question 4.
While solving the above illustrative example, can we take the lengths of PQ and PR as 5 and 13? If so, then what changes are needed In the writing of the solution. (Tcxtbook pg. no. 111)
Solution:
Yes, we can take lengths of PQ and PR as 5 and 13.
In that case, we will have to take k = 1 and solve the problem accordingly.

Question 5.
Verify that the equation ‘sin² θ + cos² θ = 1’ is true when θ = 0° or θ = 90°.
(Textbook pg. no. 112)
Solution:
sin² θ + cos² θ = 1
i. lf θ = 0°,
LH.S. = sin² θ + cos² θ
= sin² 0° + cos² 0°
= 0 + 1 …[∵ sin 0° = 0, cos 0° = 1]
= R.H.S.
∴ sin² θ + cos² θ = 1

ii. If θ = 90°,
L.H.S.= sin² θ +cos² θ
= sin² 90° + cos² 90°
= 1 + 0 … [ ∵ sin 90° = 1, cos 90° = 0]
= 1
= R.H.S.
∴ sin² θ + cos² θ = 1

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Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 8.2- Trigonometry

Download PDF: Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 8.2- Trigonometry PDF

Chapterwise Maharashtra Board Solutions Class 9 Maths :

Part 2

FAQs

Where do I get the Maharashtra State Board Books PDF For free download?

You can download the Maharashtra State Board Books from the eBalbharti official website, i.e. cart.ebalbharati.in or from this article.

How to Download Maharashtra State Board Books?

Students can get the Maharashtra Books for primary, secondary, and senior secondary classes from here. You can view or download the Maharashtra State Board Books from this page or from the official website for free of cost. Students can follow the detailed steps below to visit the official website and download the e-books for all subjects or a specific subject in different mediums.
Step 1: Visit the official website ebalbharati.in
Step 2: On the top of the screen, select “Download PDF textbooks”
Step 3: From the “Classes” section, select your class.
Step 4: From “Medium”, select the medium suitable to you.
Step 5: All Maharashtra board books for your class will now be displayed on the right side.
Step 6: Click on the “Download” option to download the PDF book.

Who developed the Maharashtra State board books?

As of now, the MSCERT and Balbharti are responsible for the syllabus and textbooks of Classes 1 to 8, while Classes 9 and 10 are under the Maharashtra State Board of Secondary and Higher Secondary Education (MSBSHSE).

How many state boards are there in Maharashtra?

The Maharashtra State Board of Secondary & Higher Secondary Education, conducts the HSC and SSC Examinations in the state of Maharashtra through its nine Divisional Boards located at Pune, Mumbai, Aurangabad, Nasik, Kolhapur, Amravati, Latur, Nagpur and Ratnagiri.

About Maharashtra State Board (MSBSHSE)

The Maharashtra State Board of Secondary and Higher Secondary Education or MSBSHSE (Marathi: महाराष्ट्र राज्य माध्यमिक आणि उच्च माध्यमिक शिक्षण मंडळ), is an autonomous and statutory body established in 1965. The board was amended in the year 1977 under the provisions of the Maharashtra Act No. 41 of 1965.

The Maharashtra State Board of Secondary & Higher Secondary Education (MSBSHSE), Pune is an independent body of the Maharashtra Government. There are more than 1.4 million students that appear in the examination every year. The Maha State Board conducts the board examination twice a year. This board conducts the examination for SSC and HSC.

The Maharashtra government established the Maharashtra State Bureau of Textbook Production and Curriculum Research, also commonly referred to as Ebalbharati, in 1967 to take up the responsibility of providing quality textbooks to students from all classes studying under the Maharashtra State Board. MSBHSE prepares and updates the curriculum to provide holistic development for students. It is designed to tackle the difficulty in understanding the concepts with simple language with simple illustrations. Every year around 10 lakh students are enrolled in schools that are affiliated with the Maharashtra State Board.