Height & Distance Test-1

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Height & Distance Test-1

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Question 1
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A tower PQ stands on a horizontal ground with base Q on the ground. The point R divides the tower in two parts such that QR = 15 m. If from a point A on the ground the angle of elevation of R is 60° and the part PR of the tower subtends an angle of 15° at A, then the height of the tower is :

A
5(2√3 + 3) m

B
5(√3 + 3) m

C
10(√3 + 1) m

D
10(2√3 + 1) m

Solution

Calculation:

From ΔAPQ
Question 2
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Two buildings are standing opposite to each other on either side of the road. From a point on the road, ladder reaches building's window which is 12 m high from the ground. Keeping the foot of the ladder at the same point, the ladder is turned on the other side of the road and it reaches a window which is 9 m high from the ground. If the length of the ladder is 15 m, then the width of the road (in m) is :

A
18

B
24

C
21

D
25

Solution

Concept Used:

In mathematical notation, the Pythagorean theorem can be expressed as:

c² = a² + b²

where c is the length of the hypotenuse and a and b are the lengths of the other two sides

Calculation:

Let's denote the distance between the foot of the ladder and the point on the road as x and with other point on the road as y.

When ladder is placed on first building

In triangle OAP, using Pythagoras Theorem

x² + 12² = 15²
⇒ x2 + 144 = 225
⇒ x2 = 81
⇒ x = 9 (as -9 can't be the value of x)

W hen the ladder is placed against the second building,

In triangle, OQC

y² + 9^2 = 15^2
⇒ y2 + 81 = 225
⇒ y2 = 144
⇒ y = 12 (as -12 can't be the value of y)

Since, x + y represents the distance between both buildings.

Thus, width of road is 12+9 = 21 m.
Question 3
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The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 20 seconds at the speed of 432 km/hour, the angle of elevation changes to 30°. If the jet plane is flying at a constant height, then its height is :

A
3600√3 m

B
2400√3 m

C
1800√3 m

D
1200√3 m

Solution

Explanation:

Given that jet plane at the initial stage at point A with an angle 60° then after 20 seconds with speed 432 km/hours angle changes to 30°

To find height first it is needed to find the distance by using speed = distance/time
Question 4
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Consider a regular hexagon ABCDEF. There are two towers placed at B and D. The angle of elevation from A to the top of the tower at B is 30°, and to the top of the tower at D is 45°. What is the ratio of the heights of towers at B and D?

A
1 : √3

B
1 : 2√3

C
1 : 2

D
3 : 4√3

Solution

Given:

A regular hexagon ABCDEF and two towers placed at B and D.

Angle of elevation from A to the top of the tower at B is 30°.

Angle of elevation from A to the top of the tower at D is 45°.

Concept Used:

Each angle of a regular hexagon is equal to 120°.

There are diagonals of two different lengths in a regular hexagon. The shortest is √3a and the longest is 2a, where a is the length of a side of the hexagon.

Calculation:

Let AB = a and BM and DN be the towers at B and D respectively.

In ΔABM, BM = AB tan 30° = a/√3.

In ΔADN, DN = AD tan 45° = 2a.

Required ratio is BM : DN = a/√3 : 2a = 1 : 2√3.
Question 5
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A man wishes to find the height of the building which stands on a horizontal plane. At a point on the plane, he finds the angle of elevation of the top of the building to be 60° and moving 10 m along the same line away from the building he finds the angle of elevation to be 30°. What is the height of the building?

A
3√3

B
5√3

C
10√3

D
12√3

Solution

Given:

The angle of elevation of 60° and 30° respectively.

Concept:

tanθ = (Perpendicular)/(Base)

Calculation:

⇒ Let AB be the height of the building.

⇒ From ΔABC, tan60° = (AB)/(BC)

⇒ BC = h/√3 ....(1)

⇒ From ΔABD, tan30° = (AB)/(BD)

⇒ (AB)/(BC + CD) = 1/√3

⇒ √3(AB) = BC + 10

⇒ BC = √3h - 10 ....(2)

⇒ From (1) and (2)

⇒ h/√3 = √3h - 10

⇒ h = 3h - 10√3

⇒ 2h = 10√3

⇒ h = 5√3

∴ The required result will be 5√3.
Question 6
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On a plane area there are two vertical towers separated by 100 feet apart. The shorter tower is 40 feet tall. A pole of length 6 feet stands on the line joining the base of two towers so that the tip of the towers and tip of the pole are also on the same line. If the distance of the pole from the shorter tower is 75 feet, then what is the height of the taller tower (approximately)?

A
85 feet

B
110 feet

C
125 feet

D
140 feet

Solution

Given:

Two vertical towers distance =100 feet

The distance of the pole from the shorter tower = 75

The shorter tower length = 40 feet

A pole length = 6 feet

Formula Used:

tan θ = Perpendicular/Base

Calculation:

In triangle FGH:

GH = 40 – 6 = 34 feet

HF = DE = 75 feet

So, tanx = GH/HF = 34/75

In triangle ABF:

BF = CE = 100 + 75 = 175 feet

So, tanx = AB/BF = AB/175

Hence,

34/75 = AB/175

⇒ AB = 7/3 × 34 = 79.33 feet

∴ Height of taller tower = AC = 79.33 + 6 = 85.33 or 85 feet (Approximately)
Question 7
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From a point A on the ground, the angle of elevation of the top of the building is 60°. If the distance between the point and building is 100m, then

Find the height of the building?

A
100√3 m

B
200√3 m

C
50√3 m

D
150√3 m

Solution

Given :

Angle of elevation of the top of the building is 60°

Distance from point A to B is 100m.

Formula used :

tan θ = Perpendicular/base ----- (1)

Calculations :

From point A the angle of elevation is 60°. So

Using equation (1) at point A, we get

⇒ tan 60° = BC/AB

⇒ √3 = BC/100

⇒ BC = 100√3 m

∴ The height of the building is 100√3 m.
Question 8
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From a point A on the ground, the angle of elevation of the top of the building is 60°. If the distance between the point and building is 100m, then

If a string is attached from point A to the top of building then find the length of the string?

A
200 m

B
400 m

C
420 m

D
450 m

Solution

Given :

Angle of elevation of the top of the building is 60°

Distance from point A to B is 100m.

Formula used :

tan θ = Perpendicular/base

Pythagoras theorem, hypotenuse² = perpendicular² + base²

Calculations :

From point A the angle of elevation is 60°. So

We know that, tan θ = Perpendicular/base

⇒ tan 60° = BC/AB

⇒ √3 = BC/100

⇒ BC = 100√3 m

Using Pythagoras theorem, we get

⇒ AC² = BC² + AB²

⇒ AC² = (100√3)² + (100)²

⇒ AC² = 30000 + 10000

⇒ AC² = 40000

⇒ AC = 200 m.

∴ The length of the string is 200m.
Question 9
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ABC is a triangular park with AB = AC = 100 metres. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at A and B are cot⁻¹(3√2) and cosec⁻¹(2√2) respectively, then the height of the tower (in meters) is:

A

B
10√5

C
20

D
25

Solution

Given,

Angle of elevation of the top of the tower at A = cot⁻¹(3√2)

⇒ cot A = 3√2

Angle of elevation of the top of the tower at B = cosec⁻¹(2√2)

⇒ cosec B = 2√2

Let the height of the vertical tower situated at the midpoint of BC be h.

In ∆ALM:

In ∆ABM, by using Pythagoras theorem,

AM²+ MB²= AB²

⇒ AM²+ MB²= (100)²

⇒ 18h²+ 7h²= 100 × 100

⇒ 25h²= 100 × 100

⇒ h²= 4 × 100

∴ h = 20 m
Question 10
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There are two buildings separated by a certain distance. The angle of elevation from the top of the small building towards the top of the long building is 30°. The angle of elevation from point A on the ground in between two buildings towards the top of the large building is 60°. The difference between the heights of the two buildings is 450 units. Find the height of the small building if the distance between the base of the small building and Point A is 150√3 units.

A
250 units

B
275 units

C
300 units

D
450 units

Solution

Given,

Let MN and BC be the heights of the large and the small buildings respectively.

⇒ NP = 450

⇒ tan30 = NP/CP

⇒ CP = MB = 450√3 units

⇒ AB = 150√3 units

⇒ AM = 450√3 – 150√3 = 300√3 units

Then,

⇒ tan60 = MN/AM

⇒ MN = 300√3 × √3

⇒ MN = 900 units

Height of the small building = BC = PM

= MN – NP = 900 – 450 = 450 units