Trigonometry
- The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 4.6 metre away from the wall. The length of the ladder is
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AB = Height of the wall
AC = Length of ladder = h metre
BC = b = 4.6 metre
∠ACB = 60°∴ cos 60° = BC AC ⇒ 1 = 4.6 2 h
⇒ h = (2 × 4.6) metre
= 9.2 metre
Correct Option: C
AB = Height of the wall
AC = Length of ladder = h metre
BC = b = 4.6 metre
∠ACB = 60°∴ cos 60° = BC AC ⇒ 1 = 4.6 2 h
⇒ h = (2 × 4.6) metre
= 9.2 metre
- A person observes that the angle of elevation at the top of a pole of height 5 metre is 30°. Then the distance of the person from the pole is :
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AB = pole = 5 metre
∠ACB = 30°, BC = ?
In ∆ABC,tan 30° = AB ⇒ 1 = 5 BC √3 BC
⇒ BC = 5 √3 metreCorrect Option: A
AB = pole = 5 metre
∠ACB = 30°, BC = ?
In ∆ABC,tan 30° = AB ⇒ 1 = 5 BC √3 BC
⇒ BC = 5 √3 metre
- A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height h. At a point on the plane, the angle of elevation of the bottom of the flag staff is a and that of the top of the flag staff is β. Then the height of the tower is
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Let height of tower = BC = y metre
AB = height of flag-staff = h metre
∠BDC = a; ∠ADC = b
Let, CD = x metre
In ∆BCD,tan α = BC CD ⇒ tan α = y ..... (i) x
In ∆ACD,tan β = AC CD ⇒ tan β = h + y x ⇒ x = h + y ..... (ii) tanβ ∴ y = h + y tanα tanβ
⇒ y tan β = h tanα + y tanα
⇒ y tanβ – y tanα = h tanα
⇒ y (tanβ – tanα) = h tanα⇒ y = h tan α tanβ - tanα
Correct Option: B
Let height of tower = BC = y metre
AB = height of flag-staff = h metre
∠BDC = a; ∠ADC = b
Let, CD = x metre
In ∆BCD,tan α = BC CD ⇒ tan α = y ..... (i) x
In ∆ACD,tan β = AC CD ⇒ tan β = h + y x ⇒ x = h + y ..... (ii) tanβ ∴ y = h + y tanα tanβ
⇒ y tan β = h tanα + y tanα
⇒ y tanβ – y tanα = h tanα
⇒ y (tanβ – tanα) = h tanα⇒ y = h tan α tanβ - tanα
- The angle of elevation of an aeroplane from a point on the ground is 45°. After flying for 15 seconds, the elevation changes to 30°. If the aeroplane is flying at a height of 2500 metres, then the speed of the aeroplane in km/ hr. is
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Let A and C be the positions of plane.
AB = CD = 2500 metre
BD = AC = x metre (let)
∠AOB = 60° ; ∠COD = 30°
In ∆OAB,
⇒ OB = 2500 metre
In ∆OCD,tan30° = CD OD ⇒ 1 √3 = 2500 2500 + x
⇒ 2500 + x = 2500√3
⇒ x
= 2500 √3 – 2500
= 2500 (√3 - 1) metre
Time = 15 seconds= 15 hour = 1 hour 60 × 60 240 ∴ Speed of plane = 2500 (√3 - 1) × 240 kmph 1000
= 600 (√3 - 1) kmph.Correct Option: D
Let A and C be the positions of plane.
AB = CD = 2500 metre
BD = AC = x metre (let)
∠AOB = 60° ; ∠COD = 30°
In ∆OAB,
⇒ OB = 2500 metre
In ∆OCD,tan30° = CD OD ⇒ 1 √3 = 2500 2500 + x
⇒ 2500 + x = 2500√3
⇒ x
= 2500 √3 – 2500
= 2500 (√3 - 1) metre
Time = 15 seconds= 15 hour = 1 hour 60 × 60 240 ∴ Speed of plane = 2500 (√3 - 1) × 240 kmph 1000
= 600 (√3 - 1) kmph.
- The thread of a kite makes angle 60° with the horizontal plane. If the length of the thread be 80 m, then the vertical height of the kite will be
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AB = Length of thread
= h metre
= 80 metre
∠BAC = 60°
BC = Vertical height of kite
In ∠ABC,sin 60° = BC AB ⇒ √3 = h 2 80 ⇒ h = 80 × √3 = 40√3 metre 2
Correct Option: D
AB = Length of thread
= h metre
= 80 metre
∠BAC = 60°
BC = Vertical height of kite
In ∠ABC,sin 60° = BC AB ⇒ √3 = h 2 80 ⇒ h = 80 × √3 = 40√3 metre 2
