Samacheer Class 9 Maths Trigonometry Book Back Answers & Solutions

Your Progress - Chapter 60% complete

1Exercise 6.2 - Trigonometric Ratios of Some Special Angles4 questions

Q.1Verify the following equalitiesv

Solution

(i) sin^2 60° + cos^2 60° = (√3/2)^2 + (1/2)^2 = 3/4 + 1/4 = 1.
(ii) 1 + tan^2 30° = 1 + (1/√3)^2 = 1 + 1/3 = 4/3; sec^2 30° = (2/√3)^2 = 4/3.
(iii) cos 90° = 0; 1 - 2 sin^2 45° = 1 - 2(1/√2)^2 = 1 - 1 = 0; 2 cos^2 45° - 1 = 2(1/2) - 1 = 0.
(iv) sin30° cos60° + cos30° sin60° = (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1 = sin90°.

Answer:

(i) 1
(ii) 4/3
(iii) 0
(iv) 1

Q.2Find the value of the followingv

Solution

(i) sin45° cos45° + sin60° cos30° = (1/√2)(1/√2) + (√3/2)(√3/2) = 1/2 + 3/4 = 5/4.
(ii) (sin90° + cos60° + cos45°)(sin30° + cos0° - cos45°) = (1 + 1/2 + 1/√2)(1/2 + 1 - 1/√2). Let a = 3/2, b = 1/√2 → (a+b)(a-b)=a^2 - b^2 = (3/2)^2 - (1/√2)^2 = 9/4 - 1/2 = 7/4.
(iii) sin^2 30° - 2 cos^3 60° + 3 tan^4 45° = (1/2)^2 - 2(1/2)^3 + 3(1)^4 = 1/4 - 1/4 + 3 = 3.

Answer:

(i) 5/4
(ii) 7/4
(iii) 3

Q.3Verify cos 3A = 4 cos^3 A - 3 cos A for A = 30°v

Solution

LHS: cos 3A = cos 90° = 0.
RHS: 4(√3/2)^3 - 3(√3/2) = 4(3√3/8) - 3√3/2 = 3√3/2 - 3√3/2 = 0.

Answer:

Verified (both sides = 0)

Q.4Find the value of 8 sin 2x cos 4x sin 6x when x = 15°v

Solution

2x = 30°, 4x = 60°, 6x = 90°. So expression = 8 sin30° cos60° sin90° = 8(1/2)(1/2)(1) = 8·1/4 = 2.

Answer:

2Exercise 6.3 - Trigonometric Ratios for Complementary Angles1 questions

Q.1Find the value of the followingv

Solution

(i) sin60° cos30° + cos60° sin30° = sin(60°+30°) = sin90° = 1.
(ii) sin(90°-30°) = cos30° = √3/2.
(iii) cos(90°-45°) = sin45° = 1/√2.
(iv) tan(90°-60°) = cot60° = 1/√3.
(v) sec(90°-30°) = cosec30° = 2.
(vi) cosec(90°-60°) = sec60° = 2.
Minimum and maximum values: for both sin and cos the range is [-1,1].

Answer:

(i) 1
(ii) √3/2
(iii) 1/√2
(iv) 1/√3
(v) 2
(vi) 2
Min/max of sin θ: -1 and 1
Min/max of cos θ: -1 and 1

3Exercise 6.4 - Trigonometric Tables and Applications6 questions

Q.1Find the value of the following (use tables)v

Solution

Values taken from standard trigonometric tables:
sin 49° ≈ 0.7547; cos 74°39' ≈ 0.2647; tan 54°26' ≈ 1.4010; sin 21°21' ≈ 0.3642; cos 33°53' ≈ 0.8300; tan 70°17' ≈ 2.7948.

Answer:

(i) 0.7547
(ii) 0.2647
(iii) 1.4010
(iv) 0.3642
(v) 0.8300
(vi) 2.7948

Q.2Find the value of θ (use tables)v

Solution

(i) sin θ = 0.9975 ⇒ θ ≈ 86°.
(ii) cos θ = 0.6763 ⇒ θ ≈ 47°25'.
(iii) tan θ = 0.0720 ⇒ θ ≈ 4°07'.
(iv) cos θ = 0.0410 ⇒ θ ≈ 87°39'.
(v) tan θ = 7.5958 ⇒ θ ≈ 82°30'.
(All angles read from standard tables/antilogs as in answerHint.)

Answer:

(i) 86°
(ii) 47°25'
(iii) 4°07'
(iv) 87°39'
(v) 82°30'

Q.3Find the value of the following (use tables)v

Solution

(i) sin 65°39' ≈ 0.9115, cos 24°57' ≈ 0.9067, tan 10°10' ≈ 0.1794 → sum = 0.9115 + 0.9067 + 0.1794 = 1.9976.
(ii) tan 70°58' ≈ 2.9042, cos 15°26' ≈ 0.9639, sin 84°59' ≈ 0.9962 → 2.9042 + 0.9639 - 0.9962 = 2.8719.

Answer:

(i) 1.9976
(ii) 2.8719

Q.4Find the area of a right triangle whose hypotenuse is 10 cm and one acute angle is 24°24'v

Solution

Hypotenuse = 10 cm. Let perpendicular p = 10 sin 24°24' ≈ 10(0.4131) = 4.131 cm. Base b = 10 cos 24°24' ≈ 10(0.9107) = 9.107 cm. Area = 1/2 · p · b ≈ 0.5·4.131·9.107 ≈ 18.81 cm^2.

Answer:

18.81 cm^2 (approx)

Q.5Find the angle made by a ladder of length 5 m with the ground, if one end is 4 m away from the wall.v

Solution

Hypotenuse = 5 m, adjacent = 4 m. cos θ = 4/5 = 0.8 ⇒ θ = arccos(0.8) ≈ 36°52'.

Answer:

36°52' (approx)

Q.6Height of tree problemv

Solution

Distance = 60 m, angle of elevation = 42°. tan 42° ≈ 0.9004 ⇒ h = 60·tan42° ≈ 60·0.9004 = 54.024 → 54.02 m (rounded).

Answer:

54.02 m (approx)

4Exercise 6.5 - Multiple Choice Questions10 questions

Q.1If sin 30° = x and cos 60° = y then x^2 + y^2 isv

Solution

sin 30° = 1/2 = x, cos 60° = 1/2 = y ⇒ x^2 + y^2 = (1/2)^2 + (1/2)^2 = 1/4 + 1/4 = 1/2.

Answer:

(1) 1/2

Q.2If tan θ = cot 37° then θ isv

Solution

cot 37° = tan(90° - 37°) = tan 53°, so θ = 53° (principal value).

Answer:

(2) 53°

Q.3Value of tan 72° · tan 18°v

Solution

72° = 90° - 18° ⇒ tan 72° = cot 18°. Thus tan72°·tan18° = cot18°·tan18° = 1.

Answer:

(2) 1

Q.4Value of (2 tan 30°) / (1 - tan^2 30°)v

Solution

Use double-angle identity: tan 2A = (2 tan A)/(1 - tan^2 A). With A = 30°, this equals tan 60°.

Answer:

(3) tan 60°

Q.5If 2 sin 2θ = √3, then θ = ?v

Solution

2 sin 2θ = √3 ⇒ sin 2θ = √3/2. So 2θ = 60° (principal value) ⇒ θ = 30°. (General solutions: 2θ = 60°+360k or 120°+360k ⇒ θ = 30°+180k or 60°+180k.)

Answer:

(2) 30°

Q.6Find the value of 3 sin 70° sec 20° + 2 sin 49° sec 51°.v

Solution

The first term is 3 sin70° sec20° = 3 because sin70° = cos20°. As printed, the second term is 2 sin49°/cos51° = 2 sin49°/sin39°, which is approximately 2.3984908. The printed total is therefore approximately 5.3984908, not one of the choices. The keyed answer 5 would be valid if the second sine were 39°, because sin39° = cos51°. This is a verified textbook misprint, not an extraction error.

Answer:

As printed: approximately 5.39849, so none of the options is correct. If sin 49° is the textbook typo and sin 39° was intended, the answer is (3) 5.

Q.7Value of (1 - tan^2 45°) / (1 + tan^2 45°) ?v

Solution

tan 45° = 1 ⇒ (1 - 1^2)/(1 + 1^2) = (1 - 1)/(1 + 1) = 0/2 = 0.

Answer:

(3) 0

Q.8Value of cosec(70°+θ) - sec(20°-θ) + tan(65°+θ) - cot(25°-θ) ?v

Solution

Use complementary identities: cosec(70°+θ) = 1/sin(70°+θ) and sec(20°-θ)=1/cos(20°-θ). Since sin(70°+θ)=cos(20°-θ), cosec(70°+θ)=sec(20°-θ). Also tan(65°+θ) = cot(25°-θ). Hence terms cancel pairwise, sum = 0.

Answer:

(1) 0

Q.9Value of tan1°·tan2°·tan3°·…·tan89° ?v

Solution

Pair terms tan A · tan(90°−A) = tan A · cot A = 1. Pairs (1°,89°), (2°,88°), … multiply to 1. Middle term tan45° = 1. Product = 1.

Answer:

(2) 1

Q.10Given sin α = 1/2 and cos β = 1/2, find α + β.v

Solution

sin α = 1/2 ⇒ α = 30° (principal). cos β = 1/2 ⇒ β = 60° (principal). So α + β = 30° + 60° = 90°.

Answer:

(2) 90°

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