Trigonometry
- There are two vertical posts, one on each side of a road, just opposite to each other. One post is108 metre high. From the top of this post, the angle of depression of the top and foot of the other post are 30° and 60° respectively. The height of the other post (in metre) is
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[ Note : Interior alternate angles
are equal ]
AB = 108 m
CD = x metre
From ∆ ABC,tan 60° = AB BC ⇒ √3 = 108 BC ⇒ BC = 108 = 36√3 m √3
From ∆ AED,tan 30° = AE ED ⇒ 1 = 108 - x √3 36√3
⇒ 108 - x = 36
⇒ 108 - x = 36 = 72 mCorrect Option: B
[ Note : Interior alternate angles
are equal ]
AB = 108 m
CD = x metre
From ∆ ABC,tan 60° = AB BC ⇒ √3 = 108 BC ⇒ BC = 108 = 36√3 m √3
From ∆ AED,tan 30° = AE ED ⇒ 1 = 108 - x √3 36√3
⇒ 108 - x = 36
⇒ 108 - x = 36 = 72 m
- If x = a cosθ + b sinθ and y = b cosθ – a sinθ, then x² + y² is equal to
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x = a cosθ + b sinθ .... (i)
y = b cosθ – a sinθ ... (ii)
On squaring both equations and adding.
x² + y² = a² cos² θ + b² sin²θ + 2ab sin θ . cos θ + b² cos² θ + a² sin² θ - 2ab sin θ . cosθ
2ab sinθ . cosθ + b² cos² + a² sin²θ - 2ab sinθ . cosθ
= a² cos²θ + a² sin²θ + b² sin²θ + b² cos²θ
= a² (cos²θ + sin²θ) + b² (sin²θ + cos²θ)
= a² + b² [∵ cos²θ + sin²θ]Correct Option: B
x = a cosθ + b sinθ .... (i)
y = b cosθ – a sinθ ... (ii)
On squaring both equations and adding.
x² + y² = a² cos² θ + b² sin²θ + 2ab sin θ . cos θ + b² cos² θ + a² sin² θ - 2ab sin θ . cosθ
2ab sinθ . cosθ + b² cos² + a² sin²θ - 2ab sinθ . cosθ
= a² cos²θ + a² sin²θ + b² sin²θ + b² cos²θ
= a² (cos²θ + sin²θ) + b² (sin²θ + cos²θ)
= a² + b² [∵ cos²θ + sin²θ]
- If (l + sin α) (l + sinβ) (l + sin γ) = (1 – sin α)(l – sinβ) (l – sin γ), then each side is equal to
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(1 + sin α) (1 + sin β) (1 + sin
γ) = (1 – sin α) (1–sin β) (1– sin γ) = x
∴ x.x = (1 + sin α) (1 – sin α) (1 + sin β) (1 – sin β)(1 + sin γ)(1– sinγ)
= (1 – sin²α) (1 – sin²β)(1 – sin²γ)
= cos²α . cos²β . cos²γ
∴ x = ± cos α . cos β . cos γCorrect Option: A
(1 + sin α) (1 + sin β) (1 + sin
γ) = (1 – sin α) (1–sin β) (1– sin γ) = x
∴ x.x = (1 + sin α) (1 – sin α) (1 + sin β) (1 – sin β)(1 + sin γ)(1– sinγ)
= (1 – sin²α) (1 – sin²β)(1 – sin²γ)
= cos²α . cos²β . cos²γ
∴ x = ± cos α . cos β . cos γ
- If r sin θ = 1, r cos θ = √3 , then the value of ( √3 tanθ + 1 ) is
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r sinθ = 1
r cosθ = √3⇒ sinθ = tanθ = 1 cosθ √3
= √3tanθ + 1= √3 × 1 + 1 = 1 + 1 = 2 √3
Correct Option: D
r sinθ = 1
r cosθ = √3⇒ sinθ = tanθ = 1 cosθ √3
= √3tanθ + 1= √3 × 1 + 1 = 1 + 1 = 2 √3
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The value of sin A + sin A is (0° < A < 90°) 1 + cos A 1 - cos A
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sinA + sinA 1 + cosA 1 - cosA = sinA(1 - cos A) + sinA (1 + cos A) = sinA – sinA cos A (1 - cos A) (1 + cos A) = + sinA + sinA . cosA 1– cos²A = 2sin A = 2 cosecA sin²A
Correct Option: A
sinA + sinA 1 + cosA 1 - cosA = sinA(1 - cos A) + sinA (1 + cos A) = sinA – sinA cos A (1 - cos A) (1 + cos A) = + sinA + sinA . cosA 1– cos²A = 2sin A = 2 cosecA sin²A
