Friday, August 24, 2012

CBSE Class 10 - Maths - CH8 - Introduction to Trigonometry

Introduction to Trigonometry
(MCQs asked in CBSE Papers)
Hipparchus: Father of Trignometry
Image credits:wikipedia

Q1: If cot A=12/5, then the value of  (sin A + cos A) × cosec A is :
(a) 13/5
(b) 17/5
(c) 14/5
(d) 1

Q2: cos 1° × cos 2° × cos 3° × ... × cos 180° is equal to :
(a)1
(b) 0
(c) 1/2
(d) –1





Q3: If sin θ = cos θ, then value of θ is
(a) 0°
(b) 45°
(c) 30°
(d) 90°

Q4: Which of the following is correct some θ such that 0° ≤ θ < 90°
(a) 1/secθ > 1
(b) 1/secθ < 1
(c)  secθ = 0
(d) 1/cosθ < 1

Q5: sin θ = 4/3 for some angle θ, is :(a) true
(b) false
(c) may be true for certain cases
(d) none of these

Q6: If sin θ = cos θ, then the value of cosec θ is :
(a) 2
(b) 1
(c) 2/√3
(d) √2

Q7: The value of sin18°/cos72° is :(a) 1
(b) 0
(c) –1
(d) 1/2

Q8: Given that sin A=1/2 and cos B=1/√2 then the value of (A + B) is:
 (a) 30°
(b) 45°
(c) 75°
(d) 15°

Q9: If sec θ + tan θ = x, then tan θ is:
(a) (x2 + 1)/x
(b) (x2 - 1)/x
(c) (x2 + 1)/2x
(d) (x2 - 1)/2x

Q10: If sec θ – tan θ = 1/3, the value of (sec θ + tan θ)
(a) 1
(b) 2
(c) 3
(d) 4

Q11: 9sec2 θ - 9tan2 θ  is equal to :
(a) 1
(b) –1
(c) 9
(d) –9

Q12: The value of tan 1°.tan 2°.tan 3° ... tan 89° is :
(a) 0
(b) 1
(c) 2
(d) ½

Q13: The value of [(cos A/cot A)  + sin A] is:
(a) cot A
(b) 2 sin A
(c) 2 cos A
(d) sec A

Q14: If tan 2A = cot (A – 18°), then the value of A is:
(a) 18°
(b) 36°
(c) 24°
(d) 27°

Q15: if 5 tan θ = 4, then value of (5 sin θ - 4 cos θ)/(5 sin θ + 4 cos θ) is:
(a) 5/3
(b) 0
(c) 5/6
(d) 1/6

Q16: If cos 3θ = √3/2; 0 < θ < 20°, then the value of  θ is :
(a) 15°
(b) 10°
(c) 0°
(d) 12°

Q17: In sin 3θ = cos (θ – 26°), where 3θ and (θ – 26°) are acute angles, then value of θ is :
(a) 30°
(b) 29°
(c) 27°
(d) 26°

Q18: sin (60° + θ) – cos (30° – θ) is equal to :
(a) 2 cos θ
(b) 2 sin θ
(c) 0
(d) 1


Answer:
1:(b) 17/5
[Hint: (sin A + cos A) × cosec A = 1 + cot A = 1 + 12/5 = 17/5]

2. (b) 0   (cos 90° = 0)

3. (b) 45°

4. (b) 1/secθ < 1    [Hint: 1/secθ = cosθ. And value of cosθ ranges from 0 to 1)]

5. (b) false [Hint: sin θ ranges from 0 to 1. It can't be greater than 1]

6. (d) √2  [Hint: cosec 45° = √2]

7.  (a) [Hint: sin18°/cos72° = sin(90°-72°)/cos72°=cos72°/cos72° = 1]

8. (c) 75° [Hint:sin A = sin 30 = 1/2 ⇒ A = 30° and cos B = cos 45° = 1/√2 ⇒ B = 45°]

9. (d) (x2 - 1)/2x
[Hint: Given sec θ + tan θ = x.               ...(I)
Also sec2 θ - tan2 θ = 1
⇒ (sec θ + tan θ)(sec θ - tan θ) = 1
⇒ sec θ - tan θ = 1/x                 ...(II)
Divide II from I,
sec θ + tan θ - (sec θ - tan θ) = x -1/x
⇒ 2tan θ = (x2 - 1)/x
⇒ tan θ = (x2 - 1)/2x]

10. (c) 3   [Hint: Use sec2 θ - tan2 θ = 1]

11: (c) 9  [Hint: Use sec2 θ - tan2 θ = 1]

12: (b) 1
[Hint: tan θ. cot θ = 1, tan(90 - θ) = cot θ  and tan45° = 1
 Given  tan 1°.tan 2°.tan 3° ... tan 88°. tan 89°
 = (tan 1°. tan 89°).(tan 2°.tan 88°) ... (tan44°.tan46°)(tan45°)
 =  [tan 1°. tan(90°-1°)].[tan2°.tan(90°-2)] ... [(tan44°.tan(90-44)].1
 =  (tan 1°. cot 1°).(tan 2°.cot2°) ... (tan44°.cot44°)
 = 1.1 ... 1 = 1]
 
13: (b) 2 sin A [Hint: Use cot A = cos A/sin A]

14: (b) 36°
[Hint: Given, tan 2A = cot (A – 18°)
⇒ tan 2A = tan(90 - (A – 18°))
⇒ tan 2A =  tan (108° – A)
⇒ 2A = 108° – A
⇒ 3A = 108°
⇒ A = 36°]

15:  (b) 0
[Hint:  Divide both numerator and denominator by cos θ and solve]

16: (b) 10° [Hint: cos 30° = √3/2]

17: (b) 29°
[Hint: sin 3θ = cos (θ – 26°)
⇒ sin 3θ = sin {90°– (θ – 26°)}
⇒ 3θ= 90° – θ+ 26° ⇒ θ= 116°/4
⇒ θ = 29°]

18: (c) 0
[Hint: sin (90° – α) = cos α
sin(60° + θ) – cos (30° – θ) = cos (90 -(60° + θ)) - cos (30° – θ) = cos (30° – θ) - cos (30° – θ) = 0]

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