CRE1 - Height & Distance | Geometry - Trigonometry
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There are two, men and a tower in between. The distance between the men is 100 m. The angle of elevation to the top of the tower are 30° and 60° respectively. Find the height of tower.
- (a)
25
- (b)
25√3
- (c)
25/√3
- (d)
None of these
Answer: Option B
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Explanation :
Given, AB = 100. Let AD = x, hence, BD = 100 – x.
In ∆ACD, tan60° = √3 =CD/AD
⇒ CD = √3 × AD …(1)
In ∆BCD, tan30° = 1/√3 = CD/BD.
⇒ CD = BD/√3 …(2)
From (1) and (2)
⇒ √3 × AD = BD/√3
⇒ 3x = 100 – x
⇒ x = 25.
∴ CD = √3 × AD = 25√3.
Hence, option (b).
Workspace:
A man is standing at some distance from a building. There is a pole at the top of the building. The building subtends an angle of 30° at the man and the angle of elevation of the top of the pole is 45°. Find the height of the pole if the building is 50m high.
- (a)
50(√3 - 1)
- (b)
50√3
- (c)
50/√3
- (d)
None of these
Answer: Option A
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Explanation :
Given, BC = 50 m.
Let CD = x.
In ∆ ABC, tan45° = 1/√3 = BC/AB
⇒ AB = BC × √3 = 50√3.
In ∆ ABD, tan45° = 1 = BD/AB
⇒ BD = AB
⇒ 50 + x = 50√3
⇒ x = 50(√3 - 1)
Hence, option (a).
Workspace:
A plane flies above a man at a height of 1000m. The angle of elevation changes from 60° to 30° in 5 sec. Find speed of plane.
- (a)
400(√3 – 1) m/s
- (b)
400√3 m/s
- (c)
400/√3 m/s
- (d)
400/(√3 + 1) m/s
Answer: Option C
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Explanation :
Given, BC = DE = 1000 m
In ∆ABC, tan60° = √3 = BC/AB
⇒ AB = BC/√3 = 1000/√3
In ∆ABC, tan30° = 1/√3 = DE/AD
⇒ AD = DE × √3 = 1000√3
∴ BD = AD - AB = 1000√3 – 1000/√3 = 2000/√3.
∴ Plane travels 2000/√3 meters in 5 seconds.
⇒ Speed of the plane = (2000/√3) ÷ 5 = 400/√3 m/s.
Hence, option (c).
Workspace:
A tree breaks down at some point and touches the ground with the tip making 30° angle with the ground. If the tree is 30 m long; find how high from the ground it broke (in meters).
- (a)
12
- (b)
15
- (c)
20
- (d)
10
Answer: Option D
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Explanation :
Let the tree broke at a height of x meters from the ground i.e., BC = x.
∴ AC = 30 – x.
In ∆ABC, sin30° = 1/2 = BC/AC
⇒ AC = 2BC
⇒ 30 – x = 2x
⇒ x = 10 meters.
Hence, option (d).
Workspace:
A ladder makes 60° with ground when it leans against the wall. The ladder slides down by 5 m along the ground making the angle 45°. Find the length of the ladder.
- (a)
10√2
- (b)
10/(√2 + 1)
- (c)
10/(√2 - 1)
- (d)
10(√2 + 1)
Answer: Option D
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Explanation :
Let the length of the ladder be x, i.e., AC = DE = x.
In ∆ABC, cos60° = 1/2 = AB/AC
⇒ AB = AC/2 …(1)
In ∆DBE, cos45° = 1/√2 = DB/DE
⇒ DB = DE/√2 …(2)
We know, DB – AB = 5.
From (1) and (2)
DE/√2 – AC/2 = 5
⇒ x/√2 – x/2 = 5
⇒ (√2 - 1)x/2 = 5
⇒ x = 10/(√2 - 1) = 10(√2 + 1).
Hence, option (d).
Workspace:
Harsha is riding vertically in a hot air balloon, directly over a point A on the ground. Harsha spots a parked car on the ground at an angle of depression of 30°. The balloon rises 50 meters. Now the angle of depression to the car is 60 degrees. How far is the car from point A?
- (a)
25
- (b)
25√3
- (c)
25/√3
- (d)
None of these
Answer: Option B
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Explanation :
Given, BD = 50 m.
Let AC = x.
In ∆CAB, tan30° = 1/√3 = AB/AC
⇒ AB = x/√3 = x/√3 …(1)
In ∆CAD, tan60° = √3 = AD/AC
⇒ AD = √3 × x = √3 × x …(2)
We know, AD – AB = BD = 50
From (1) and (2)
⇒ √3x – x/√3 = 50
⇒ 2x/√3 = 50
⇒ x = 25√3.
Hence, option (b).
Workspace:
If the shadow of a building increases by 10 meters when the angle of elevation of the sun rays decreases from 60° to 45°, what is the height of the building?
- (a)
10√3
- (b)
10/(√3 - 1)
- (c)
10√3/(√3 + 1)
- (d)
None of these
Answer: Option B
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Explanation :
Initially the shadow of CB is AB and later the shadow becomes DB.
Let BC = x
In ∆ABC, tan60° = √3 = BC/AB
⇒ AB = x/√3
In ∆DBC, tan45° = 1 = BC/DB
⇒ DB = x
We know, DB – AB = AD = 10 meters.
⇒ x – x/√3 = 10
⇒ (√3 - 1)x/√3 = 10
⇒ x = 10√3/(√3 - 1)
Hence, option (b).
Workspace:
From the top of a 200 meters high building, the angle of depression to the bottom of a second building is 60 degrees. From the same point, the angle of elevation to the top of the second building is 45 degrees. Calculate the height of the second building.
- (a)
200(√3 + 1)/√3
- (b)
200(√3 + 1)
- (c)
200√3
- (d)
None of these
Answer: Option A
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Explanation :
Given, AB = 200 meters. Let CD = x.
∴ DE = 200 and CE = x – 200.
In ∆AED, tan60° = √3 = DE/AE
⇒ AE = DE/√3 …(1)
In ∆AEC, tan45° = 1 = CE/AE
⇒ AE = CE …(2)
From (1) and (2)
DE/√3 = CE
⇒ 200/√3 = x – 200
⇒ x = 200/√3 + 200 = 200(√3 + 1)/√3
Hence, option (a).
Workspace:
A plane flies above a man at a height of 1000 m. At one time, angle of elevation is 60°. Suddenly, the plane climbs up at an angle of 30° and 5 seconds later angle of elevation becomes 45°. Speed of plane is?
- (a)
200√3
- (b)
2000/√3
- (c)
400√3
- (d)
400/√3
Answer: Option D
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Explanation :
Given, height of plane initially (BC) = 1000 meters.
Let CF = x
In ∆CFE, cos30° = √3/2 = CE/CF
⇒ CE = √3x/2 = BD …(1)
Also, sin30° = 1/2 = FE/CF
⇒ FE = x/2 …(2)
In ∆ABC, tan60° = √3 = CB/AB
⇒ AB = CB/√3 = 1000/√3 …(3)
In ∆AFD, tan45° = 1 = FD/AD
⇒ AD = FD …(4)
We know, BD = AD – AB = FD – AB
⇒ BD = (DE + EF) - AB
From (1), (2) and (3)
√3x/2 = (1000 + x/2) – 1000/√3
Solving this we get x = 2000/√3
∴ Plane travels 2000/√3 meters in 5 seconds.
⇒ Speed of the plane = (2000/√3) ÷ 5 = 400/√3.
Hence, option (d).
Workspace:
A man looks up at a pole making angle of elevation is 45°. When he looks at the reflection of the tip of the pole in water, he finds the angle of depression as 60°. If the man is 1 m tall, find height of the pole.
- (a)
1/(3 - 2√3)
- (b)
2 - √3
- (c)
2 + √3
- (d)
None of these
Answer: Option C
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Explanation :
EA is the line representing the ground.
Height of the man (represented by AB) = 1 meter.
Let height of the pole (ED) = x meters.
In ∆BFD, tan45° = 1 = DF/BF
⇒ BF = DF …(1)
In ∆BFD, tan60° = √3 = CF/BF
⇒ BF = CF/√3 …(2)
From (1) and (2)
DF = CF/√3
⇒ x – 1 = (x + 1)/√3
⇒ √3x - √3 = x + 1
⇒ √3x - x = 1 + √3
⇒ x = (√3 + 1)/(√3 - 1)
⇒ x = (√3 + 1)2/2
⇒ x = (4 + 2√3)/2 = 2 + √3
Hence, option (c).
Workspace:
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