Mathematics Multiple Choice Questions & Answers on “Trigonometric Ratios – 1”.
1. If sin (A + B) = (frac {sqrt {3}}{2}) and tan (A – B) = 1. What are the values of A and B?
a) 37, 54
b) 35.7, 40.7
c) 50, 10
d) 52.5, 7.5
Answer: d
Clarification: The value of sin (A + B) = (frac {sqrt {3}}{2}) and sin 60° = (frac {sqrt {3}}{2})
∴ A + B = 60 (1)
The value of tan (A – B) = 1 and tan 45° = 1
∴ A – B = 45 (2)
Adding equation (1) and (2)
A + B = 60
+ A – B = 45
– – – – – – – – – – – – –
2 A = 105
A = 52.5
∴ B = 7.5
2. If cos θ = (frac {3}{4}) then value of cos 2θ is ___________
a) (frac {1}{6})
b) (frac {1}{4})
c) (frac {1}{8})
d) (frac {3}{8})
Answer: c
Clarification: cos 2θ = 2cos θ2 – 1
cos θ = (frac {3}{4})
cos 2θ = 2((frac {3}{4}))2 – 1
= (frac {1}{8})
3. If sin A = (frac {8}{17}), what will be the value of cos A sec A?
a) 2
b) -1
c) 1
d) 0
Answer: c
Clarification: sin A = (frac {8}{17})
cos A sec A can be written as cosA × (frac {1}{secA}) = 1
∴ cos A sec A = 1
4. The value of each of the trigonometric ratios of an angle depends on the size of the triangle and does not depend on the angle.
a) True
b) False
Answer: b
Clarification:
Consider, two triangles ABC and DEF
In ∆ABC,
sin B = (frac {AC}{AB} = frac {10}{20} = frac {1}{2}) i.e. B = 30°
Now, in ∆DEF,
sin F = (frac {DE}{DF} = frac {20}{40} = frac {1}{2}) i.e. F = 30°
From these examples it is evident that the value of the trigonometric ratios depends on their angle and not on their lengths.
5. If tan α = √3 and cosec β = 1, then the value of α – β?
a) -30°
b) 30°
c) 90°
d) 60°
Answer: a
Clarification: tan α = √3 and tan 60° = √3
∴ α = 60°
Cosec β = 1 and cosec 90° = 1
∴ β = 90°
α – β = 60 – 90 = -30°
6. In triangle ABC, right angled at C, then the value of cosec (A + B) is __________
a) 2
b) 0
c) 1
d) ∞
Answer: c
Clarification:
Since the triangle is right angles at C,
The sum of the remaining two angles will be 90
∴ cosec(A + B) = Cosec 90° = 1
7. If tan θ = (frac {3}{4}) then the value of sinθ is _________
a) (frac {3}{5})
b) (frac {4}{4})
c) (frac {3}{4})
d) (frac {-3}{5})
Answer: a
Clarification:
tanθ = (frac {BC}{AC} = frac {3}{4} = frac {3k}{4k})
Hence, BC = 3k, AC = 4k
Using Pythagoras theorem
AB2 = AC2 + BC2
AB2 = 4k2 + 3k2
AB = 5k
sinθ = (frac {BC}{AB} = frac {3k}{5k} = frac {3}{5})