# CRE 5 - Trigonometry | Geometry, Mensuration & Trigonometry

**CRE 5 - Trigonometry | Geometry, Mensuration & Trigonometry**

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

**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).

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**CRE 5 - Trigonometry | Geometry, Mensuration & Trigonometry**

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

**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).

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**CRE 5 - Trigonometry | Geometry, Mensuration & Trigonometry**

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

**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).

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**CRE 5 - Trigonometry | Geometry, Mensuration & Trigonometry**

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

**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).

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**CRE 5 - Trigonometry | Geometry, Mensuration & Trigonometry**

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

**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).

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**CRE 5 - Trigonometry | Geometry, Mensuration & Trigonometry**

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

**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).

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**CRE 5 - Trigonometry | Geometry, Mensuration & Trigonometry**

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

**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).

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**CRE 5 - Trigonometry | Geometry, Mensuration & Trigonometry**

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

**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).

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**CRE 5 - Trigonometry | Geometry, Mensuration & Trigonometry**

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

**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:

**CRE 5 - Trigonometry | Geometry, Mensuration & Trigonometry**

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

**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).

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