Trigonometry
- A kite is flying at the height of 75 m from the ground. The string makes an angle q (where cotq = 8/15 ) with the level ground. Assuming that there is no slack in the string the length of the string is equal to :
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A = Position of kite
AC = length of string
AB = 75 metrecot θ = 8 15
∴ cosecθ = √1 + cot²θ= √ + 1 8 ² = √ + 1 64 15 225 = √ 225 + 64 = √ + 1 289 225 225 = 17 15 ∴ sinθ = 17 15
From ∆ABCsinθ = AB AC ⇒ 15 = 75 17 AC
⇒ AC × 15 = 17 × 15⇒ AC = 17 × 75 = 85 metre 15 Correct Option: A
A = Position of kite
AC = length of string
AB = 75 metrecot θ = 8 15
∴ cosecθ = √1 + cot²θ= √ + 1 8 ² = √ + 1 64 15 225 = √ 225 + 64 = √ + 1 289 225 225 = 17 15 ∴ sinθ = 17 15
From ∆ABCsinθ = AB AC ⇒ 15 = 75 17 AC
⇒ AC × 15 = 17 × 15⇒ AC = 17 × 75 = 85 metre 15
- The angle of elevation of a tower from a distance of 100 metre from its foot is 30°. Then the height of the tower is
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AB = Tower = h metre
From ∆ABC,tan 30° = AB BC ⇒ 1 = h √3 100 >⇒ h = 100 metre √3 Correct Option: D
AB = Tower = h metre
From ∆ABC,tan 30° = AB BC ⇒ 1 = h √3 100 >⇒ h = 100 metre √3
- A 10 metre long ladder is placed against a wall. It is inclined at an angle of 30° to the ground. The distance (in m) of the foot of the ladder from the wall is (Given √3 = 1.732)
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AC = ladder = 10 metre
BC = ?
∠ABC = θ = 30°
From ∆ABC,cos θ = BC AC ⇒ cos 30° = BC 10 ⇒ √3 = BC 2 10 ⇒ BC = 10√3 = 5√3 2
= 5 × 1.732 = 8.660 metreCorrect Option: D
AC = ladder = 10 metre
BC = ?
∠ABC = θ = 30°
From ∆ABC,cos θ = BC AC ⇒ cos 30° = BC 10 ⇒ √3 = BC 2 10 ⇒ BC = 10√3 = 5√3 2
= 5 × 1.732 = 8.660 metre
- If the height of a pole is 2 √3 metre and the length of its shadow is 2 metre, then the angle of elevation of the sun is
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tan θ = AB = 2√3 = √3 BC 2
⇒ tan θ = tan 60° ⇒ θ = 60°Correct Option: D
tan θ = AB = 2√3 = √3 BC 2
⇒ tan θ = tan 60° ⇒ θ = 60°
- From two points on the ground and lying on a straight line through the foot of a pillar, the two angles of elevation of the top of the pillar are complementary to each other. If the distances of the two points from the foot of the pillar are 12 metres and 27 metres and the two points lie on the same side of the pillar, then the height (in metres) of the pillar is
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Let, ∠ACB = θ
∴ ∠ADB = 90° – θ
BC = 12 metre,
BD = 27 metre
AB = Pillar = h metre
From ∆ABC,tan θ = AB = h ......(i) BC 12
From ∆ABDtan(90° – θ) = AB BD ⇒ cotθ = h ......(ii) 27 ∴ tanθ. cotθ = h × h 12 27
⇒ h2 = 12 × 27
⇒ h2 = √12 × 27
= √2 × 2 × 3 × 3 × 3 × 3
= 2 × 3 × 3 = 18 metreCorrect Option: B
Let, ∠ACB = θ
∴ ∠ADB = 90° – θ
BC = 12 metre,
BD = 27 metre
AB = Pillar = h metre
From ∆ABC,tan θ = AB = h ......(i) BC 12
From ∆ABDtan(90° – θ) = AB BD ⇒ cotθ = h ......(ii) 27 ∴ tanθ. cotθ = h × h 12 27
⇒ h2 = 12 × 27
⇒ h2 = √12 × 27
= √2 × 2 × 3 × 3 × 3 × 3
= 2 × 3 × 3 = 18 metre