Trigonometry
- The angles of depression of two ships from the top of a light house are 45° and 30° toward east. If the ships are 200m apart, the height of the light house is (Take √3 =1.73)
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AB = Height of light-post = h metre
CD = 200 metre;
C and D ⇒ positions of ships
∠ACB = 45°; ∠ADB = 30°
In ∆ABC,tan 45° = AB BC ⇒ 1 = AB ⇒ AB = BC = h metre BC
In ∆ABD,tan 30° = AB BD ⇒ 1 = h √3 h + 200
⇒ √3h = h + 200
⇒ √3h – h = 200
⇒ h (√3 - 1 ) = 200⇒ h = 200 √3
- 1= 200(√3
+ 1)(√3
- 1)(√3
+ 1)= 200(√3
+ 1)2
= 100 (1.73 + 1) metre
= 273 metreCorrect Option: A
AB = Height of light-post = h metre
CD = 200 metre;
C and D ⇒ positions of ships
∠ACB = 45°; ∠ADB = 30°
In ∆ABC,tan 45° = AB BC ⇒ 1 = AB ⇒ AB = BC = h metre BC
In ∆ABD,tan 30° = AB BD ⇒ 1 = h √3 h + 200
⇒ √3h = h + 200
⇒ √3h – h = 200
⇒ h (√3 - 1 ) = 200⇒ h = 200 √3
- 1= 200(√3
+ 1)(√3
- 1)(√3
+ 1)= 200(√3
+ 1)2
= 100 (1.73 + 1) metre
= 273 metre
- A helicopter, at an altitude of 1500 metre, finds that two ships are sailing towards it, in the same direction. The angles of depression of the ships as observed from the helicopter are 60° and 30° respectively. Distance between the two ships, in metre is
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AD = Height of helicopter = 1500 metre
C and D ⇒ positions of ships
∠ADB = 30°; ∠ACB = 60°
Let, BC = x metre and BD = y metre
In ∆ABD,tan 30° = AB BD ⇒ 1 = 1500 √3 y
⇒ y = 1500 √3 metre ...(i)
In ∆ABC,tan 60° = AB BC ⇒√3
=1500 x ⇒x = 1500 √3
= 500 √3 metre ...(ii)
∴ Distance between ships
= (y – x) metre
= (1500 √3 500 √3 - ) metre
= 1000 √3 metreCorrect Option: A
AD = Height of helicopter = 1500 metre
C and D ⇒ positions of ships
∠ADB = 30°; ∠ACB = 60°
Let, BC = x metre and BD = y metre
In ∆ABD,tan 30° = AB BD ⇒ 1 = 1500 √3 y
⇒ y = 1500 √3 metre ...(i)
In ∆ABC,tan 60° = AB BC ⇒√3
=1500 x ⇒x = 1500 √3
= 500 √3 metre ...(ii)
∴ Distance between ships
= (y – x) metre
= (1500 √3 500 √3 - ) metre
= 1000 √3 metre
- From the top of a tower 60 metre high the angle of depression of the top and bottom of a pole are observed to be 45° and 60° respectively. If the pole and tower stand on the same plane, the height of the pole in metre is
-
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AB = height of tower = 60 metre
BE = CD = height of pole
= h metre
BC = ED = x metre
In ∆ AED,tan 45° = AE ED ⇒ 1
=60 - h x
60 – h ... (i)
In ∆ ABC,tan 60° = AB BC ⇒√3
=60 x
⇒ √3 x = 60⇒x
=60 = 20√3 metre √3
From equation (i),
20 √3 = 60 – h
⇒ h = 60 - 20√3
= 20 (3 - √3) metreCorrect Option: C
AB = height of tower = 60 metre
BE = CD = height of pole
= h metre
BC = ED = x metre
In ∆ AED,tan 45° = AE ED ⇒ 1
=60 - h x
60 – h ... (i)
In ∆ ABC,tan 60° = AB BC ⇒√3
=60 x
⇒ √3 x = 60⇒x
=60 = 20√3 metre √3
From equation (i),
20 √3 = 60 – h
⇒ h = 60 - 20√3
= 20 (3 - √3) metre
- The cliff of a mountain is 180 m high and the angles of depression of two ships on the either side of cliff are 30° and 60°. What is the distance between the two ships?
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AD = Cliff = 180 metre
∠ABD = 60°, ∠ACD = 30°
From ∆ABD,tan 60° = AD BD ⇒ √3
=180 BD ⇒BD
=180 = 60√3
metre√3
From ∆ACD,tan 30° = AD CD ⇒ 1 = 180 √3 CD
⇒ CD = 100 √3 metre
∴ BC = BD + DC
= 60 √3 + 180 √3
= 240 √3 metre
= (240 × 1.732) metre
= 415.68 metreCorrect Option: C
AD = Cliff = 180 metre
∠ABD = 60°, ∠ACD = 30°
From ∆ABD,tan 60° = AD BD ⇒ √3
=180 BD ⇒BD
=180 = 60√3
metre√3
From ∆ACD,tan 30° = AD CD ⇒ 1 = 180 √3 CD
⇒ CD = 100 √3 metre
∴ BC = BD + DC
= 60 √3 + 180 √3
= 240 √3 metre
= (240 × 1.732) metre
= 415.68 metre
- Two posts are 2 metres apart. Both posts are on same side of a tree. If the angles of depressions of these posts when observed from the top of the tree are 45° and 60° respectively, then the height of the tree is :
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CD = 2 metre
BD = x metre
AB = Tree = h metre
From ∆ ABC,tan 45° = AB BC ⇒ 1 = h x + 2
⇒ h = (x + 2) metre .....(i)
From ∆ ABD,tan 60° = AB BD ⇒ √3
=h x ⇒ x
=h ....(ii) √3
From equations (i) and (ii),h = h + 2 √3 ⇒ h - h = 2 √3 ⇒ √3
h - h= 2 √3
⇒ h(√3 - 1) = 2 √3⇒ h = 2√3 √3
- 1= 2√3(√3
+ 1)(√3 - 1)(√3 + 1) = 2√3(√3
+ 1)= √3
(√3
+ 1)3 - 1
= (3 + √3) metreCorrect Option: B
CD = 2 metre
BD = x metre
AB = Tree = h metre
From ∆ ABC,tan 45° = AB BC ⇒ 1 = h x + 2
⇒ h = (x + 2) metre .....(i)
From ∆ ABD,tan 60° = AB BD ⇒ √3
=h x ⇒ x
=h ....(ii) √3
From equations (i) and (ii),h = h + 2 √3 ⇒ h - h = 2 √3 ⇒ √3
h - h= 2 √3
⇒ h(√3 - 1) = 2 √3⇒ h = 2√3 √3
- 1= 2√3(√3
+ 1)(√3 - 1)(√3 + 1) = 2√3(√3
+ 1)= √3
(√3
+ 1)3 - 1
= (3 + √3) metre
