Trigonometry
- If the angle of elevation of a cloud from a point 200m above a lake is 30° and the angle of depression of its reflection in the lake is 60°. Then the height of the cloud above the lake is :
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AB is the surface of lake. C’ is the reflection of cloud ‘C’.
∠CPM = 30° and ∠C'PM = 60°
Let, C M = h metre
CB = (h + 200) metre
C'B = (h + 200) metre
In ∆CMP,tan 30° = CM PM ⇒ 1 = h √3 PM
⇒ PM = √3 h ....(i)
In ∆PMC',tan 60° = C'M PM ⇒ tan 60° = C'B + BM PM ⇒ √3 = h + 200 + 200 PM ⇒ PM = h + 400 .... (ii) √3
From equations (i) and (ii) ,√3 h = h + 400 √3
⇒ 3h = h + 400
⇒ 2h = 400 ⇒ h = 200
∴ CB = h + 200 = 400 metre
Note : If the angle of elevation of a cloud from a point h metre above a lake is a and the angle of depression of its reflection in the lake is b, thenThe height of the cloud = h( tanβ + tanα ) ( tanβ - tanα )
Correct Option: D
AB is the surface of lake. C’ is the reflection of cloud ‘C’.
∠CPM = 30° and ∠C'PM = 60°
Let, C M = h metre
CB = (h + 200) metre
C'B = (h + 200) metre
In ∆CMP,tan 30° = CM PM ⇒ 1 = h √3 PM
⇒ PM = √3 h ....(i)
In ∆PMC',tan 60° = C'M PM ⇒ tan 60° = C'B + BM PM ⇒ √3 = h + 200 + 200 PM ⇒ PM = h + 400 .... (ii) √3
From equations (i) and (ii) ,√3 h = h + 400 √3
⇒ 3h = h + 400
⇒ 2h = 400 ⇒ h = 200
∴ CB = h + 200 = 400 metre
Note : If the angle of elevation of a cloud from a point h metre above a lake is a and the angle of depression of its reflection in the lake is b, thenThe height of the cloud = h( tanβ + tanα ) ( tanβ - tanα )
- The shadow of a tower when the angle of elevation of the sun is 45°, is found to be 10 metre longer than when it was 60°. The height of the tower is
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AB = Height of tower
= h metre
BC = Length of shadow when ∠BCA = 60° = x metre
BD = Length of shadow when ∠ADB = 45° = (x + 10) metre
In ∆ABC,tan 60° = AB BC ⇒ √3 = h x
⇒ h = 3 x metre ..... (i)
In ∆ABD,tan 45° = AB ⇒ 1 = h BD x + 10 ⇒ h = x + 10 ⇒ h = h + 10 √3 ⇒ h - h = 10 √3 ⇒ √3h - h = 10 √3
⇒ h(√3 - 1) = 10√3⇒ h = 10√3 = 10√3(√3 + 1) √3 - 1 (√3 - 1)( √3 + 1)
= 5(3 + √3) metreCorrect Option: B
AB = Height of tower
= h metre
BC = Length of shadow when ∠BCA = 60° = x metre
BD = Length of shadow when ∠ADB = 45° = (x + 10) metre
In ∆ABC,tan 60° = AB BC ⇒ √3 = h x
⇒ h = 3 x metre ..... (i)
In ∆ABD,tan 45° = AB ⇒ 1 = h BD x + 10 ⇒ h = x + 10 ⇒ h = h + 10 √3 ⇒ h - h = 10 √3 ⇒ √3h - h = 10 √3
⇒ h(√3 - 1) = 10√3⇒ h = 10√3 = 10√3(√3 + 1) √3 - 1 (√3 - 1)( √3 + 1)
= 5(3 + √3) metre
- The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. The height of the tower is :
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Let, AB = Height of tower = h metre
BC = 4 metre, BD = 9 metre
∠ACB = 90° – θ; ∠ADB = θ
In ∆ABC,tan (90° – θ) = AB BC ⇒ cotθ = h ..........(i) 4
In ∆ABD,tanθ = h ..........(ii) 9
On multiplying both equations,tanθ.cotθ = h × h 4 9 h² = 1 ⇒ h² = 36 36
⇒ √36 = 6 metreCorrect Option: D
Let, AB = Height of tower = h metre
BC = 4 metre, BD = 9 metre
∠ACB = 90° – θ; ∠ADB = θ
In ∆ABC,tan (90° – θ) = AB BC ⇒ cotθ = h ..........(i) 4
In ∆ABD,tanθ = h ..........(ii) 9
On multiplying both equations,tanθ.cotθ = h × h 4 9 h² = 1 ⇒ h² = 36 36
⇒ √36 = 6 metre
- A hydrogen-filled balloon ascending at the rate of 18 kmph was drifted by wind. Its angle of elevation at 10th and 15th minutes were found to be 60° and 45° respectively. The wind speed (in whole numbers) during the last five minutes, approximately, is equal to
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If height of balloon = 3800 √3 m, then.tan 60° = BP AB ⇒ √3 = 3800√3 AB
⇒ AB = 3800 mtan 45° = CQ AC ⇒ 1 = 3800√3 AC
⇒ AC = 3800 √3 m
∴ PQ = AC – AB
= (3800 √3 – 3800)
= 3800 × 0.732 = 3800 ( √3 –1)
= 2782 m
&thete4; Required speed= 
2782 
km/hr. (5/60) × 1000
= 33.3 km/hr. (Approximately)Correct Option: D
If height of balloon = 3800 √3 m, then.tan 60° = BP AB ⇒ √3 = 3800√3 AB
⇒ AB = 3800 mtan 45° = CQ AC ⇒ 1 = 3800√3 AC
⇒ AC = 3800 √3 m
∴ PQ = AC – AB
= (3800 √3 – 3800)
= 3800 × 0.732 = 3800 ( √3 –1)
= 2782 m
&thete4; Required speed= 2782 km/hr. (5/60) × 1000
= 33.3 km/hr. (Approximately)
- The distance between two pillars is 120 metres. The height of one pillar is thrice the other. The angles of elevation of their tops from the mid point of the line connecting their feet are complementary to each other. The height (in metres) of the taller pillar is (Use : √3 = 1.732)
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Let, AB = 3h metre
CD = h metre
BE = ED = 60° metre
∠AEB = 90° – θ ; ∠CED = θ
In ∆ABE,tan (90° – θ) = AB BE ⇒ cot θ = 3h .......(i) 60
In ∆CED,tan θ = h .......(ii) 60 ∴ tanθ . cotθ = 3h × h 60 60
⇒ 3h2 = 60 × 60⇒ h2 = 60 × 60 = 1200 3
⇒ h = √1200
= 20 √3 metre
∴ Height of larger pole
= 3 × 20 √3 = 60 √3 metre
= (60 × 1.732) metre
= 103.92 metreCorrect Option: D
Let, AB = 3h metre
CD = h metre
BE = ED = 60° metre
∠AEB = 90° – θ ; ∠CED = θ
In ∆ABE,tan (90° – θ) = AB BE ⇒ cot θ = 3h .......(i) 60
In ∆CED,tan θ = h .......(ii) 60 ∴ tanθ . cotθ = 3h × h 60 60
⇒ 3h2 = 60 × 60⇒ h2 = 60 × 60 = 1200 3
⇒ h = √1200
= 20 √3 metre
∴ Height of larger pole
= 3 × 20 √3 = 60 √3 metre
= (60 × 1.732) metre
= 103.92 metre
