Trigonometry
- The top of a broken tree touches the ground at a distance of 15 metre from its base. If the tree is broken at a height of 8 metre from the ground, then the actual height of the tree is
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Let AB = height of tree = h metre
AC = 8 metre,
BC = CD = Broken part of tree
AD = 15 metre
In ∆ACD,
CD² = AC² + AD² = 8² + 15²
= 64 + 225 = √289
∴ CD = 289 = 17 metre
∴ Original height of tree
= 17 + 8 = 25 metreCorrect Option: C
Let AB = height of tree = h metre
AC = 8 metre,
BC = CD = Broken part of tree
AD = 15 metre
In ∆ACD,
CD² = AC² + AD² = 8² + 15²
= 64 + 225 = √289
∴ CD = 289 = 17 metre
∴ Original height of tree
= 17 + 8 = 25 metre
- A straight tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle of 30° with the ground. The distance from the foot of the tree to the point, where the top touches the ground is 10 m. Find the total height of the tree?
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AB = Height of tree
BD = 10 metre
AC = CD = broken part of tree
∠CDB = 30°
In ∆BCDtan30° BC BD ⇒ 1 = BC √3 10 ⇒ BC = 10 metre √3 Again, sin 30° = BC CD ⇒ 1 = 10 2 √3 × CD ⇒ CD = 10 × 2 = 20 metre √3 √3
∴ AB = BC + CD= 10 + 20 = 30 √3 √3 √3
= 10 √3 metreCorrect Option: A
AB = Height of tree
BD = 10 metre
AC = CD = broken part of tree
∠CDB = 30°
In ∆BCDtan30° BC BD ⇒ 1 = BC √3 10 ⇒ BC = 10 metre √3 Again, sin 30° = BC CD ⇒ 1 = 10 2 √3 × CD ⇒ CD = 10 × 2 = 20 metre √3 √3
∴ AB = BC + CD= 10 + 20 = 30 √3 √3 √3
= 10 √3 metre
- A 1.6 m tall observer is 45 metres away from a tower. The angle of elevation from his eye to the top of the tower is 30°, then the height of the tower in metres is (Take √3 = 1.732)
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AB = Height of observer = 1.6 metre
CD = Height of tower = h metre
∴ DE = (h – 1.6) metre ; BC = AE = 45 metre
∠ DAE = 30°∴ tan30° DE AE ⇒ 1 = h - 1.6 √3 45 ⇒ h - 1.6 = 45 = 15√3 √3
⇒ h – 1.6 = 15 × 1.732 = 25.98
⇒ h = (25.98 + 1.6) metre
= 27.58 metreCorrect Option: C
AB = Height of observer = 1.6 metre
CD = Height of tower = h metre
∴ DE = (h – 1.6) metre ; BC = AE = 45 metre
∠ DAE = 30°∴ tan30° DE AE ⇒ 1 = h - 1.6 √3 45 ⇒ h - 1.6 = 45 = 15√3 √3
⇒ h – 1.6 = 15 × 1.732 = 25.98
⇒ h = (25.98 + 1.6) metre
= 27.58 metre
- The angle of elevation of the top of a tower from a point A on the ground is 30°. On moving a distance of 20 metres towards the foot of the tower to a point B, the angle of elevation increases to 60°. The height of the tower in metres is
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Let, AB = height of tower = h metre
∠ ACB = 30°,
∠ADB = 60°
CD = 20 metre ; BC = x metre
In ∆ ABC,tan30° = AB BC ⇒ 1 = h √3 x
⇒ x = √3 h .... (i)
In ∆ ABD,tan60° = AB BD ⇒ √3 = h x - 20
⇒ h = √3 x – 20 √3
= √3 √3 h – 20 √3
⇒ h = 3h – 20 √3
⇒ 3h – h = 20 √3
⇒ 2h = 20 √3⇒ h = 20√3 = 10√3 metre 2
Correct Option: C
Let, AB = height of tower = h metre
∠ ACB = 30°,
∠ADB = 60°
CD = 20 metre ; BC = x metre
In ∆ ABC,tan30° = AB BC ⇒ 1 = h √3 x
⇒ x = √3 h .... (i)
In ∆ ABD,tan60° = AB BD ⇒ √3 = h x - 20
⇒ h = √3 x – 20 √3
= √3 √3 h – 20 √3
⇒ h = 3h – 20 √3
⇒ 3h – h = 20 √3
⇒ 2h = 20 √3⇒ h = 20√3 = 10√3 metre 2
- A person of height 6ft. wants to pluck a fruit which is on a (26 / 3) ft. high tree. If the person is standing (8 / √3) ft. away from the base of the tree, then at what angle should he throw a stone so that it hits the fruit?
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AB = Tree = 26 feet 3
BE = CD = 6 feetAE = AB – BE = 26 - 6 3 = 26 - 18 = 8 feet 3 3 DE = BC = 8 feet √3
From ∆AEDtan θ = AE = (8 / 3) ED (8 / √3) = 8 × √3 = 1 3 8 √3
Correct Option: B
AB = Tree = 26 feet 3
BE = CD = 6 feetAE = AB – BE = 26 - 6 3 = 26 - 18 = 8 feet 3 3 DE = BC = 8 feet √3
From ∆AEDtan θ = AE = (8 / 3) ED (8 / √3) = 8 × √3 = 1 3 8 √3
