Trigonometry
- A tower standing on a horizontal plane subtends a certain angle at a point 160 m apart from the foot of the tower. On advancing 100 m towards it, the tower is found to subtend an angle twice as before. The height of the tower is
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AB = Tower = h metre
CD = 100 metre; BC = 160 metre
∠ACB = ∴ ∠ADB = 2θ
In ∆ ABC,tanθ = AB BC ⇒ tanθ = h .............(i) 160
In ∆ ABD,tan2θ = AB BD ⇒ tan2θ = h 60 ⇒ 2tanθ = h 1 - tan2θ 60 = 2 × h 160 1 - h2 160 × 160 h 60 ⇒ 1 = 1 80 
1 - h2 
60 160 × 160 ⇒ 4 1 - h2 = 3 160 × 160 ⇒ h2 = 1 - 3 = 1 160 × 160 4 4
⇒ h2 = 6400
⇒ h = √6400 = 80 metreCorrect Option: A
AB = Tower = h metre
CD = 100 metre; BC = 160 metre
∠ACB = ∴ ∠ADB = 2θ
In ∆ ABC,tanθ = AB BC ⇒ tanθ = h .............(i) 160
In ∆ ABD,tan2θ = AB BD ⇒ tan2θ = h 60 ⇒ 2tanθ = h 1 - tan2θ 60 = 2 × h 160 1 - h2 160 × 160 h 60 ⇒ 1 = 1 80 1 - h2 60 160 × 160 ⇒ 4 1 - h2 = 3 160 × 160 ⇒ h2 = 1 - 3 = 1 160 × 160 4 4
⇒ h2 = 6400
⇒ h = √6400 = 80 metre
- A vertical stick 12 cm long casts a shadow 8 cm long on the ground. At the same time, a tower casts a shadow 40 m long on the ground. The height of the tower is
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Height of tower Length of stick = Length of shadow of tower Length of shadow of stick ⇒ h = 40 12 8 ⇒ h = 40 × 12 = 60 metre 8 Correct Option: B
Height of tower Length of stick = Length of shadow of tower Length of shadow of stick ⇒ h = 40 12 8 ⇒ h = 40 × 12 = 60 metre 8
- If the angle of elevation of a balloon from two consecutive kilometre-stones along a road are 30° and 60° respectively, then the height of the balloon above the ground will be
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AB = Height of balloon = h km
BD = x km, CD = 1 km
From ∆ ABD.tan 60° = AB BD ⇒ √3 = h x ⇒ x = h km ......(i) √3
From ∆ ABC,tan 30° = AB BC ⇒ 1 = h √3 h + 1 √3 ⇒ √3 h = h + 1 √3 ⇒ √3 h - h = 1 √3 ⇒ 3h - h = 1 √3
⇒ 2h = √3⇒ h = √3 km 2 Correct Option: A
AB = Height of balloon = h km
BD = x km, CD = 1 km
From ∆ ABD.tan 60° = AB BD ⇒ √3 = h x ⇒ x = h km ......(i) √3
From ∆ ABC,tan 30° = AB BC ⇒ 1 = h √3 h + 1 √3 ⇒ √3 h = h + 1 √3 ⇒ √3 h - h = 1 √3 ⇒ 3h - h = 1 √3
⇒ 2h = √3⇒ h = √3 km 2
- A kite is flying at a height of 50 metre. If the length of string is 100 metre then the inclination of string to the horizontal ground in degree measure is
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AB = Height of kite = 50 metre
OB = length of thread = 100 metre∴ sin BOA = AB = 50 = 1 = sin 30° OB 100 2
∴ ∠BOA = 30°Correct Option: D
AB = Height of kite = 50 metre
OB = length of thread = 100 metre∴ sin BOA = AB = 50 = 1 = sin 30° OB 100 2
∴ ∠BOA = 30°
- The angle of elevation of a tower from a distance 100 m from its foot is 30°. Height of the tower is :
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AB = Tower = h metre
∠ACB = 30°;
BC = 100 metre∴ tan30° = AB BC ⇒ 1 = h √3 100 ⇒ h = 100 metre √3 Correct Option: A
AB = Tower = h metre
∠ACB = 30°;
BC = 100 metre∴ tan30° = AB BC ⇒ 1 = h √3 100 ⇒ h = 100 metre √3
