Trigonometry
- The angles of depression of two ships from the top of a light house are 60° and 45° towards east. If the ships are 300 metre apart, the height of the light house is
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AB = Lamp post = h metre C and D = Positions of ships CD = 300 metre; BC = x metre ∆ACB = 60 metre; ∠ADB = 45°
In ∆ABC,tan 60° = AB BC ⇒ √3 = h x
⇒ h = √3x ..... (i)
In ∆ABD,tan 45° = AB BD ⇒ 1 = h x + 300
⇒ h = x + 300⇒ h = h + 300 √3 ⇒ h - h = 300 √3 ⇒ h - √3h - h = 300 √3
⇒ h (√3 - 1) = 300√3⇒ h = 300(√3) √3 - 1 = 300√3(√3 + 1) (√3 - 1)(√3 + 1) = 300(3 + √3) 2
= 150(3 + √3) metre
= 45 kmph.
Correct Option: C
AB = Lamp post = h metre C and D = Positions of ships CD = 300 metre; BC = x metre ∆ACB = 60 metre; ∠ADB = 45°
In ∆ABC,tan 60° = AB BC ⇒ √3 = h x
⇒ h = √3x ..... (i)
In ∆ABD,tan 45° = AB BD ⇒ 1 = h x + 300
⇒ h = x + 300⇒ h = h + 300 √3 ⇒ h - h = 300 √3 ⇒ h - √3h - h = 300 √3
⇒ h (√3 - 1) = 300√3⇒ h = 300(√3) √3 - 1 = 300√3(√3 + 1) (√3 - 1)(√3 + 1) = 300(3 + √3) 2
= 150(3 + √3) metre
= 45 kmph.
- The angle of elevation of an aeroplane as observed from a point 30 metre above the transparent water-surface of a lake is 30° and the angle of depression of the image of the aeroplane in the water of the lake is 60°. The height of the aeroplane from the water-surface of the lake is
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AB = transparent water-surface ∆CPM = 30°; ∠C’PM = 60°
CM = h
CB = h + 30
∴ C’B = h + 30
In ∆CMP,tan 30° = CM PM ⇒ 1 ⇒ h √3 PM
⇒ PM = 3h ..... (i)
In ∆PMC’tan 60° = C'M PM ⇒ √3 = h + 30 + 30 PM ⇒ PM = h + 60 ...(ii) √3 ∴ √3h = h + 60 √3
⇒ 3h = h + 60
⇒ 2h = 60
⇒ h = 30
∴ CB = BM + CM = 30 + 30 = 60 metreCorrect Option: A
AB = transparent water-surface ∆CPM = 30°; ∠C’PM = 60°
CM = h
CB = h + 30
∴ C’B = h + 30
In ∆CMP,tan 30° = CM PM ⇒ 1 ⇒ h √3 PM
⇒ PM = 3h ..... (i)
In ∆PMC’tan 60° = C'M PM ⇒ √3 = h + 30 + 30 PM ⇒ PM = h + 60 ...(ii) √3 ∴ √3h = h + 60 √3
⇒ 3h = h + 60
⇒ 2h = 60
⇒ h = 30
∴ CB = BM + CM = 30 + 30 = 60 metre
- A boat is moving away from an observation tower. It makes an angle of depression of 60° with an observer’s eye when at a distance of 50 metre from the tower. After 8 seconds, the angle of depression becomes 30°. By assuming that it is running in still water, the approximate speed of the boat is :
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AB = Height of observation tower = h metre
C and D = Positions of boat
BC = 50 metre
Let, CD = x metre
∠ACB = 60° = ∠EAC
∠ADB = 30° = ∠EAD
In ∆ABC,tan 60° = AB BC ⇒ √3 = h 50
⇒ h = 50√3 metre
In ∆ABD,tan 30° = AB BD ⇒ 1 = 50√3 √3 50 + x
⇒ 50 + x = 50√3 × √3 = 150
⇒ x = 150 – 50 = 100 metre∴ Speed of boat = Distance Time = 
100 
m/sec. 8 = 100 × 18 kmph 8 5
Correct Option: C
AB = Height of observation tower = h metre
C and D = Positions of boat
BC = 50 metre
Let, CD = x metre
∠ACB = 60° = ∠EAC
∠ADB = 30° = ∠EAD
In ∆ABC,tan 60° = AB BC ⇒ √3 = h 50
⇒ h = 50√3 metre
In ∆ABD,tan 30° = AB BD ⇒ 1 = 50√3 √3 50 + x
⇒ 50 + x = 50√3 × √3 = 150
⇒ x = 150 – 50 = 100 metre∴ Speed of boat = Distance Time = 100 m/sec. 8 = 100 × 18 kmph 8 5
- From a point on a bridge across the river, the angles of depression of the banks on opposite sides of the river are 30° and 45° respectively. If the bridge is at a height of 2.5 m from the banks, then the width of the river is (Take √3 = 1.732)
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AD = Height of bridge = 2.5 metre
∠EAB = ∠ABD = 45°;
∠FAC = ∠ACD = 30°
In ∆ABD,tan 45° = AD ⇒ 1 = 2.5 ⇒ CD = 2.5 metre BD BD
In ∆ACD,tan 30° = AD ⇒ 1 = 2.5 CD √3 CD
⇒ CD 2.5√3 metre
CD = (2.5 × 1.732) metre = 4.33 metre
∴ BC = (2.5 + 4.33) metre = 6.83 metre
Correct Option: B
AD = Height of bridge = 2.5 metre
∠EAB = ∠ABD = 45°;
∠FAC = ∠ACD = 30°
In ∆ABD,tan 45° = AD ⇒ 1 = 2.5 ⇒ CD = 2.5 metre BD BD
In ∆ACD,tan 30° = AD ⇒ 1 = 2.5 CD √3 CD
⇒ CD 2.5√3 metre
CD = (2.5 × 1.732) metre = 4.33 metre
∴ BC = (2.5 + 4.33) metre = 6.83 metre
- From the top of a building 60 metre high, the angles of depression of the top and bottom of a tower are observed to be 30° and 60° respectively. The height of the tower in metre is :
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AB = Height of building = 60 metre
CD = Height of tower = h metre
∠FAD = ∠ADE = 30°
∠FAC = ∠ACB = 60°
From ∆ABC,tan60° = AB BC ⇒ √3 = 60 BC ⇒ BC = 60 = 20√3 metre = DE √3
From ∆ADE,tan 30° = AE ED ⇒ 1 = 60 - h √3 20√3
⇒ h = 60 – 20 = 40 metreCorrect Option: A
AB = Height of building = 60 metre
CD = Height of tower = h metre
∠FAD = ∠ADE = 30°
∠FAC = ∠ACB = 60°
From ∆ABC,tan60° = AB BC ⇒ √3 = 60 BC ⇒ BC = 60 = 20√3 metre = DE √3
From ∆ADE,tan 30° = AE ED ⇒ 1 = 60 - h √3 20√3
⇒ h = 60 – 20 = 40 metre
