The angle of elevation of a tower from a distance 100 m from its foot is 30°. The height of the tower will be –
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Solution
⇒ AB = 100 tan 30
⇒ AB =
= 57.73 meter
If , then sin θ – cos θ = ?
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Solution
= k (Let)
sin θ = kx, cos θ = ky
⇒ sin2θ + cos2θ = 1 = k2(x2 + y2)
So,
So, sin θ – cos θ = K (x – y) =
The angle of elevation of top and bottom of a flag kept on a flag post which is 30 metre apart from a man are 45° and 30° respectively. What is the height of flag?
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Solution
tan 45° =
So AC = CD = 30 meter
tan 30° =
BC =
So, AB = AC – BC =
= 30 – 17.32
= 12.68 m
If sec 5A = cosec (A + 36°), where 5A is an acute angle. Then A = ?
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Solution
sec 5A = cosec (A + 36°)
⇒ sec 5A = sec [90° – (A + 36°)]
⇒ sec 5A = sec [54° – A]
⇒ 5A = 54° – A
= 6A = 54°
⇒ A = 9°
If 4cos2θ – 4 cos θ + 1 = 0 then tan (θ – 15) = ?
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Solution
4 cos2θ – 4 cos θ + 1 = 0
⇒ (2 cos θ – 1)2 = 0
⇒ θ = 60°
So tan (60° – 15°) = tan 45° = 1
If tan 2A = cot (A – 18) where 2A is an acute angle then A = ?
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Solution
tan 2A = cot (A – 18)
= tan[90 – (A – 18)]
= tan [108 – A]
So 2A = 108 – A
⇒ 3A = 108°
⇒ A = 36°
If x sin3θ + y cos3θ = sin θ cos θ and x sin θ = y cos θ then x2 + y2 = ?
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Solution
x sin3θ + y cos3θ = sin θ cos θ
⇒ (x sin θ) sin2θ + (y cos θ) cos2θ = sin θ cos θ
= (x sin θ) sin2θ + x sin θ cos2θ = sin θ cos θ
= x sin θ [sin2θ + cos2θ] = sin θ cos θ
⇒ x = cos θ and y = sin θ
So x2 + y2 = cos2θ + sin2θ = 1
If sin θ + sin2θ = 1 then cos2θ + cos4θ = ?
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Solution
sin θ + sin2θ = 1
⇒ sin θ = 1 – sin2θ
⇒ sin θ = cos2θ
⇒ sin2θ = cos4θ
⇒ 1 – cos2θ = cos4θ
⇒ cos4θ + cos2θ = 1
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Solution
If a cos θ – b sin θ = c, then a sin θ + b cos θ = ?
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Solution
c = a cosθ – b sinθ
c2 = (a cosθ – b sinθ)2 = a2cos2θ + b2 sin2θ – 2ab sinθ cosθ
= a2 (1 – sin2θ) + b2 (1 – cos2θ) – 2ab sinθ cosθ
= a2 + b2 – a2 sin2θ – b2cos2θ – 2ab sinθ cosθ
= a2 + b2 – (a sinθ + b cosθ)2
(a sinθ + b cosθ)2 = a2 + b2 – c2
a sinθ + b cosθ =