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Chapter 8 Trigonometry Practice Set 8.2 Solutions

Complete Maharashtra State Board Balbharati & Yuvakbharati textbook solutions for Chapter 8 Trigonometry Practice Set 8.2. Step-by-step solved exercises, numerical problems, and digest answers.

9 Solved Questions• 1562 words

Practice Set 8.2 Geometry 9th Std Maths Part 2 Answers Chapter 8 Trigonometry

Question 1 Maharashtra Board Solution
In the following table, a ratio is given in each column. Find the remaining two ratios in the column and complete the table.
Solution & Step-by-Step Answer:
i. cos θ = …(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 = 2 … [Taking square root of both sides] = 12k

ii. sin θ = …..(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 = ...[Taking square root of both sides]
= 60k

iii. tan θ = 1 = ..(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 =

iv. sin θ = ..(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 = ...[Taking square root of both sides]
=

v. cos θ = ..(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 = ...[Taking square root of both sides]
AB = √2K

vi. cos θ = ..(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 = ...[Taking square root of both sides]
= 29K

vii. tan θ = ..(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 = ...[Taking square root of both sides]
= 17K

viii. sin θ = ..(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 = ...[Taking square root of both sides]
= 4K

ix. tan θ = ..(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 = ...[Taking square root of both sides]
= 3K

Question 2 Maharashtra Board Solution
Find the values of: i. 5 sin 30° + 3 tan 45° ii. tan2 60° + 3 sin2 60° iii. 2 sin 30° + cos 0° + 3 sin 90° iv. v. cos2 45° + sin2 30° vi. cos 60° x cos 30° + sin 60° x sin 30°
Solution & Step-by-Step Answer:
i. sin 30° = and tan 45° = 1

ii. tan260° + 3 sin260°

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

iv.

v. cos245° + sin230°

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

Question 3 Maharashtra Board Solution
If sin θ = , then find cos θ.
Solution & Step-by-Step Answer:
sin θ = ...(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 = ..[Taking square root of both sides] = 3k
Question 4 Maharashtra Board Solution
If cos θ = , then find sin θ.
Solution & Step-by-Step Answer:
cos θ = ...(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 = ..[Taking square root of both sides] = 8k

Maharashtra Board Class 9 Maths Chapter 8 Trigonometry Practice Set 8.2 Intext Questions and Activities

Question 1 Maharashtra Board Solution
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 & Step-by-Step Answer:
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 Maharashtra Board Solution
In right angled ∆PQR, ∠Q = 90°, ∠R = θ and if sin θ = , then find cos θ and tan θ. (Textbook pg. no. 110)
Solution & Step-by-Step Answer:
i. Take the given trigonometric ratio as 13k equation (i). sin θ = ...(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 = ... [Taking square root of both sides] = 12k
Question 3 Maharashtra Board Solution
While solving the above Illustrative example, why the lengths of PQ and PR are taken 5k and 13k? (Textbook pg. no. 111)
Solution & Step-by-Step Answer:
= … [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 Maharashtra Board Solution
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 & Step-by-Step Answer:
(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 Maharashtra Board Solution
Verify that the equation ‘sin2 θ + cos2 θ = 1’ is true when θ = 0° or θ = 90°. (Textbook pg. no. 112)
Solution & Step-by-Step Answer:
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.= sin2θ +cos2θ
= sin290° + cos290°
= 1 + 0 … [ ∵ sin 90° = 1, cos 90° = 0]
= 1
= R.H.S.
∴ sin2θ + cos2θ = 1

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Chapter 8 Trigonometry Practice Set 8.1
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Chapter 8 Trigonometry Problem Set 8