Trigonometry
- If the angle of elevation of the sun decreases from 45° to 30°, then the length of the shadow of a pillar increases by 60m. The height of the pillar is
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AB = Height of pole = h metre
CD = 60 metre
In ∆ABCtan 45° = AB ⇒ 1 = h BC BC
⇒ BC = h metre
In ∆ABD,tan 30° = AB BD ⇒ 1 = h √3 h + 60
⇒ √3 h = h + 60
⇒ √3 h – h = 60
⇒ h (√3 - 1) = 60⇒ h = 60 √3 - 1 = 60(√3 + 1) = 60(√3 + 1) = 90 (√3 - 1)(√3 + 1) 2
= 30 (√3 + 1) metreCorrect Option: C
AB = Height of pole = h metre
CD = 60 metre
In ∆ABCtan 45° = AB ⇒ 1 = h BC BC
⇒ BC = h metre
In ∆ABD,tan 30° = AB BD ⇒ 1 = h √3 h + 60
⇒ √3 h = h + 60
⇒ √3 h – h = 60
⇒ h (√3 - 1) = 60⇒ h = 60 √3 - 1 = 60(√3 + 1) = 60(√3 + 1) = 90 (√3 - 1)(√3 + 1) 2
= 30 (√3 + 1) metre
- An aeroplane flying horizontally at a height of 3 km. above the ground is observed at a certain point on earth to subtend an angle of 60°. After 15 seconds of flight, its angle of elevation is changed to 30°. The speed of the aeroplane (Take, √3 = 1.732) is
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AB = CD = 3000 metre
A and C = Positions of aeroplane
∠AOB = 60°; ∠COD = 30°
In ∆OAB,tan 60° = AB OB ⇒√3 = 3000 OB ⇒ OB = 3000 √3
= 1000 √3 metre
In ∆OCD,tan 30° = CD OD ⇒ 1 = 3000 √3 OD
⇒ OD = 3000 √3 metre
∴ BD = (3000 √3 – 1000 √3 ) metre
= 2000 √3 metre
∴ Speed of aeroplane= 2000√3 m./sec. 15 = 
2000 × 1.732 
m./sec. 15
= 230.93 m./sec.Correct Option: B
AB = CD = 3000 metre
A and C = Positions of aeroplane
∠AOB = 60°; ∠COD = 30°
In ∆OAB,tan 60° = AB OB ⇒√3 = 3000 OB ⇒ OB = 3000 √3
= 1000 √3 metre
In ∆OCD,tan 30° = CD OD ⇒ 1 = 3000 √3 OD
⇒ OD = 3000 √3 metre
∴ BD = (3000 √3 – 1000 √3 ) metre
= 2000 √3 metre
∴ Speed of aeroplane= 2000√3 m./sec. 15 = 2000 × 1.732 m./sec. 15
= 230.93 m./sec.
- If the length of the shadow of a vertical pole be √3 times the height of the pole, the angle of elevation of the sun is :
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AB = Height of pole = x units
BC = Length of shadow
= √3 x units
∠ACB = θ∴ tan θ = AB BC ⇒ tan θ = x = 1 = tan 30° √3 x √3
⇒ θ = 30°Correct Option: C
AB = Height of pole = x units
BC = Length of shadow
= √3 x units
∠ACB = θ∴ tan θ = AB BC ⇒ tan θ = x = 1 = tan 30° √3 x √3
⇒ θ = 30°
- The angle of elevation of the top of a tower from two horizontal points (in opposite sides) at distances of 25 metre and 64 metre from the base of tower are x and 90° – x respectively. The height of the tower will be
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Let AB = Height of tower = h metre
BC = 25 metre
BD = 64 metre
∠ACB = x° and ∠ADB = (90 – x)
In ∆ABC,tan x = AB BC ⇒ tan x = h 25
In ∆ABD,tan (90° – x) = AB BD ⇒ cot x = h 64 ∴ tanx . cotx = h × h 25 64
⇒ h² = 25 × 64
∴ h = √25 × 64 = 5 × 8
= 40 metreCorrect Option: D
Let AB = Height of tower = h metre
BC = 25 metre
BD = 64 metre
∠ACB = x° and ∠ADB = (90 – x)
In ∆ABC,tan x = AB BC ⇒ tan x = h 25
In ∆ABD,tan (90° – x) = AB BD ⇒ cot x = h 64 ∴ tanx . cotx = h × h 25 64
⇒ h² = 25 × 64
∴ h = √25 × 64 = 5 × 8
= 40 metre
- From two points, lying on the same horizontal line, the angles of elevation of the top of the pillar are θ and φ (θ < φ). If the height of the pillar is ‘h’ m and the two points lie on the same sides of the piller, then the distance between the two points is
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Let AB = height of pole = h metre
∠ACB = θ, ∠ADB = φ
In ∆ABD,tan φ = AB BD ⇒ BD = h = h cot φ tan φ
In ∆ABC,tan θ = AB BC ⇒ BC = h = h cot θ tan θ
∴ Required distance
= CD = h cotθ – h cotφ
= h (cotθ – cotφ) metreCorrect Option: C
Let AB = height of pole = h metre
∠ACB = θ, ∠ADB = φ
In ∆ABD,tan φ = AB BD ⇒ BD = h = h cot φ tan φ
In ∆ABC,tan θ = AB BC ⇒ BC = h = h cot θ tan θ
∴ Required distance
= CD = h cotθ – h cotφ
= h (cotθ – cotφ) metre
