Trigonometry
- A man on the top of a tower, standing on the sea shore, finds that a boat coming towards him takes 10 minutes for the angle of depression to change from 30° to 60°. How soon the boat reach the seashore?
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Let speed of boat = v metre/minute
Time taken to reach B from D = t minutes
∠ACB = 30°; ∠ADB = 60°
AB = Tower
In ∆ABC,tan 30° = AB BC ⇒ 1 = AB √3 vt + 10v ⇒ AB = vt + 10v √3
In ∆ABD,tan 60° = AB BD ⇒ √3 = AB vt
⇒ √3 vt = AB⇒ √3vt = 10v + vt √3
⇒ 3t = 10 + t
⇒ 2t = 10
⇒ t = 5 minutesCorrect Option: A
Let speed of boat = v metre/minute
Time taken to reach B from D = t minutes
∠ACB = 30°; ∠ADB = 60°
AB = Tower
In ∆ABC,tan 30° = AB BC ⇒ 1 = AB √3 vt + 10v ⇒ AB = vt + 10v √3
In ∆ABD,tan 60° = AB BD ⇒ √3 = AB vt
⇒ √3 vt = AB⇒ √3vt = 10v + vt √3
⇒ 3t = 10 + t
⇒ 2t = 10
⇒ t = 5 minutes
- From the top of a cliff 100 metre high, the angles of depression of the top and bottom of a tower are 45° and 60° respectively. The height of the tower is
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AB = Height of cliff = 100 metre.
CD = Height of tower = h metre.
∠ADE = 45°, ∠ACB = 60°
In ∆ABC,tan 60° = AB BC ⇒ √3
=100 BC ⇒ BC = 100 metre ....(1) √3
In ∆ADE,tan 45° = AE DE ⇒ 1 = AE = 100 - h BC BC
⇒ BC = 100 – h∴ 100 = 100 - h √3 ⇒ h = 100 – 100 √3 = 100√3
- 100√3 = 100(√3 - 1) = 100√3(√3 - 1) √3 3 = 100(3 - √3) metre 3 Correct Option: A
AB = Height of cliff = 100 metre.
CD = Height of tower = h metre.
∠ADE = 45°, ∠ACB = 60°
In ∆ABC,tan 60° = AB BC ⇒ √3
=100 BC ⇒ BC = 100 metre ....(1) √3
In ∆ADE,tan 45° = AE DE ⇒ 1 = AE = 100 - h BC BC
⇒ BC = 100 – h∴ 100 = 100 - h √3 ⇒ h = 100 – 100 √3 = 100√3
- 100√3 = 100(√3 - 1) = 100√3(√3 - 1) √3 3 = 100(3 - √3) metre 3
- If secθ + tanθ = 2, then the value of sec θ is
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sec θ + tan θ = 2 ......(i)
∴ sec² θ – tan² θ = 1
⇒ (sec θ + tan θ) (sec θ – tan θ) = 1⇒ sec θ – tan θ = 1 ......(ii) 2
By adding equations (i) and (ii),
∴ sec θ + tan θ + sec θ – tan θ= 2 + 1 = 5 2 2 ⇒ 2 sec θ = 5 ⇒ sec θ = 5 2 4
Correct Option: C
sec θ + tan θ = 2 ......(i)
∴ sec² θ – tan² θ = 1
⇒ (sec θ + tan θ) (sec θ – tan θ) = 1⇒ sec θ – tan θ = 1 ......(ii) 2
By adding equations (i) and (ii),
∴ sec θ + tan θ + sec θ – tan θ= 2 + 1 = 5 2 2 ⇒ 2 sec θ = 5 ⇒ sec θ = 5 2 4
- A pilot in an aeroplane at an altitude of 200 metre observes two points lying on either side of a river. If the angles of depression of the two points be 45° and 60°, then the width of the river is
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P = Position of pilot ;
PC = 200 metre
AB = width of river
AC = x metre (let)
CB = y metre (let)
∠PAC = 45° ; ∠PBC = 60°
In ∆ APC,tan45° = PC AC ⇒ 1 = 200 x
⇒ x = 200 metre
In ∆ PCB,tan 60° = PC CB ⇒ √3
=200 y ⇒ y
=200 metre √3
∴ Width of river = x + y= 
200 + 200 
metre √3 Correct Option: A
P = Position of pilot ;
PC = 200 metre
AB = width of river
AC = x metre (let)
CB = y metre (let)
∠PAC = 45° ; ∠PBC = 60°
In ∆ APC,tan45° = PC AC ⇒ 1 = 200 x
⇒ x = 200 metre
In ∆ PCB,tan 60° = PC CB ⇒ √3
=200 y ⇒ y
=200 metre √3
∴ Width of river = x + y= 200 + 200 metre √3
- The angle of depression of a point situated at a distance of 70 m from the base of a tower is 60°. The height of the tower is :
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AB = Tower = h metre (let)
∠DAC = ∠ACB = 60°
BC = 70 metre
In ∆ABC,tan 60° = AB BC ⇒ √3
=h 70
⇒ h = 70 √3 metre
Correct Option: B
AB = Tower = h metre (let)
∠DAC = ∠ACB = 60°
BC = 70 metre
In ∆ABC,tan 60° = AB BC ⇒ √3
=h 70
⇒ h = 70 √3 metre
