Trigonometry
- From the top of a tower of height 180 m the angles of depression of two objects on either sides of the tower are 30° and 45°. Then the distance between the objects are
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AD is tower and B and C are two objects,
∠ABD = 30° and ∠ACD = 45°
AD = 180 metre
From ∆ABD,tan 30° = AD BD ⇒ 1 = 180 √3 BD
⇒ BD = 180 √3 metre
From ∆ADC,tan 45° = AD DC ⇒ 1 = 180 ⇒ DC = 180 metre DC
∴ BC = BD + DC
= 180 √3 + 180
= 180 ( √3 + 1) metreCorrect Option: D
AD is tower and B and C are two objects,
∠ABD = 30° and ∠ACD = 45°
AD = 180 metre
From ∆ABD,tan 30° = AD BD ⇒ 1 = 180 √3 BD
⇒ BD = 180 √3 metre
From ∆ADC,tan 45° = AD DC ⇒ 1 = 180 ⇒ DC = 180 metre DC
∴ BC = BD + DC
= 180 √3 + 180
= 180 ( √3 + 1) metre
- From a tower 125 metres high, the angle of depression of two objects, which are in horizontal line through the base of the tower, are 45° and 30° and they are on the same side of the tower. The distance (in metres) between the objects is
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AB = Tower = 125 metre
BC = x metre, BD = y metre
From ∆ABC,tan 45° = AB BD ⇒ 1 = 125 ⇒ x = 125 metre x
From ∆ABD,tan 30° = AB BD ⇒ 1 = 125 √3 y
⇒ y = 125 √3 metre
⇒ CD = y – x = 125 √3 – 125
= 125 ( √3 – 1) metreCorrect Option: B
AB = Tower = 125 metre
BC = x metre, BD = y metre
From ∆ABC,tan 45° = AB BD ⇒ 1 = 125 ⇒ x = 125 metre x
From ∆ABD,tan 30° = AB BD ⇒ 1 = 125 √3 y
⇒ y = 125 √3 metre
⇒ CD = y – x = 125 √3 – 125
= 125 ( √3 – 1) metre
- From the top of a hill 200 m high, the angle of depression of the top and the bottom of a tower are observed to be 30° and 60°. The height of the tower is (in m) :
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AB = Hill = 200 metre
∠ADE = 30°
∠ACB = 60°
DE = BC = x metre
From ∆ABC,tan 60° = AB BC ⇒√3 = 200 x ⇒ x = 200 metre √3
From ∆AEDtan 30° = AE DE ⇒ 1 = AE √3 (200 / √3) ⇒ AE = 200 metre 3 ∴ CD = 200 – 200 = 400 3 3 = 133 1 metre 3
Correct Option: C
AB = Hill = 200 metre
∠ADE = 30°
∠ACB = 60°
DE = BC = x metre
From ∆ABC,tan 60° = AB BC ⇒√3 = 200 x ⇒ x = 200 metre √3
From ∆AEDtan 30° = AE DE ⇒ 1 = AE √3 (200 / √3) ⇒ AE = 200 metre 3 ∴ CD = 200 – 200 = 400 3 3 = 133 1 metre 3
- The top of two poles of height 24 m and 36 m are connected by a wire. If the wire makes an angle of 60° with the horizontal, then the length of the wire is
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DE = 36 – 24 = 12 metre
From ∆ ADE,sin 60° = DE AD ⇒ √3 = 12 2 AD ⇒ AD = 12 × 2 = 8√3 metre √3
Correct Option: B
DE = 36 – 24 = 12 metre
From ∆ ADE,sin 60° = DE AD ⇒ √3 = 12 2 AD ⇒ AD = 12 × 2 = 8√3 metre √3
- There are two temples, one on each bank of a river, just opposite to each other. One temple is 54m high. From the top of this temple, the angles of depression of the top and the foot of the other temple are 30° and 60° respectively. The length of the temple is :
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AB = temple = 54 metre
CD = temple = h metre
BC = width of river
= x metre
From ∆ABC,tan 60° = AB BC ⇒√3 = 54 ⇒ x = 54 x √3
⇒ 18√3 metre
From ∆ ADEtan 30° = AE DE &rArr 1 = 54 - h √3 18√3
⇒54 – h = 18
⇒ h = 54 – 18 = 36 metreCorrect Option: B
AB = temple = 54 metre
CD = temple = h metre
BC = width of river
= x metre
From ∆ABC,tan 60° = AB BC ⇒√3 = 54 ⇒ x = 54 x √3
⇒ 18√3 metre
From ∆ ADEtan 30° = AE DE &rArr 1 = 54 - h √3 18√3
⇒54 – h = 18
⇒ h = 54 – 18 = 36 metre
