# Applications of Trigonometry

Chapters

- ap-scert-10th-class-maths-lesson-1-Real-Numbers
- ap-scert-10th-class-Maths-Lesson-2-Sets
- ap-scert-10th-class-Maths-Lesson-3-Polynomials
- ap-scert-10th-class-Maths-Lesson-4-Pair-of-Linear-Equations-in-Two-Variables
- ap-scert-10th-class-Maths-Lesson-5-Quadratic-Equations
- ap-scert-10th-class-Maths-Lesson-6-Progressions
- ap-scert-10th-class-Maths-Lesson-7-Coordinate-Geometry
- ap-scert-10th-class-Maths-Lesson-8-Similar-Triangles
- ap-scert-10th-class-Maths-Lesson-9-Tangents-and-Secants-to-a-Circle
- ap-scert-10th-class-Maths-Lesson-10-Mensuration
- ap-scert-10th-class-Maths-Lesson-11-Trigonometry
- ap-scert-10th-class-Maths-Lesson-12-Applications-of-Trigonometry
- ap-scert-10th-class-Maths-Lesson-13-Probability
- ap-scert-10th-class-Maths-Lesson-14-Statistics

- ap-scert-10th-class-maths-lesson-1-Real-Numbers
- ap-scert-10th-class-Maths-Lesson-2-Sets
- ap-scert-10th-class-Maths-Lesson-3-Polynomials
- ap-scert-10th-class-Maths-Lesson-4-Pair-of-Linear-Equations-in-Two-Variables
- ap-scert-10th-class-Maths-Lesson-5-Quadratic-Equations
- ap-scert-10th-class-Maths-Lesson-6-Progressions
- ap-scert-10th-class-Maths-Lesson-7-Coordinate-Geometry
- ap-scert-10th-class-Maths-Lesson-8-Similar-Triangles
- ap-scert-10th-class-Maths-Lesson-9-Tangents-and-Secants-to-a-Circle
- ap-scert-10th-class-Maths-Lesson-10-Mensuration
- ap-scert-10th-class-Maths-Lesson-11-Trigonometry
- ap-scert-10th-class-Maths-Lesson-12-Applications-of-Trigonometry
- ap-scert-10th-class-Maths-Lesson-13-Probability
- ap-scert-10th-class-Maths-Lesson-14-Statistics

- Solutions
- PDF Download
- Question Papers
- Videos

Solutions

A tower stands vertically on the ground. From a point which is 15 meter away from the foot of the tower, the angle of elevation of the top of the tower is 45°. What is the height of the tower?

Answer:

Let the height of the tower = h m

Distance of the point of observation from the foot of the tower =15 cm.

Angle of elevation of the top of the tower = 45°

From the figure tan θ = opp. side/adj. side

tan 45° =h/15

⇒ 1 =h/15

∴ h = 1 × 15 = 15 m

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground by making 30° angle with the ground. The distance between the foot of the tree and the top of the tree on the ground is 6 m. Find the height of the tree before falling down.

Answer:

Distance between the foot of tree and the point of contact of the top of the tree on the ground = 6 cm.

Let the length of the remaining part be = h m.

Let the length of the broken part be = x m.

Angle made by the broken part with the ground = 30°.

From the figure

tan 30° =h/6

∴ Height of the tree = broken part + remaining part

= x + h

= 2√3 + 4√3 = 6√3 m

= 6 × 1.732

≃ 10.392 m.

A contractor wants to set up a slide for the children to play in the park. He wants to set it up at the height of 2 m and by making an angle of 30° with the ground. What should be the length of the slide?

Answer:

Height of slide = 2 m

Let the length of the slide = x m.

Angle made by the slide with the ground = 30°

From the figure

sin 30° =2/x

⇒1/2=2/x

⇒ x = 2 × 2 = 4 m

Length of the slide = 4 m.

Length of the shadow of a 15 meter high pole is 5√3 meters at 7 o’clock in the morning. Then, what is the angle of elevation of the Sun rays with the ground at the time?

Answer:

Height of the pole = 15 m

Length of the shadow = 5√3 m

Let the angle of elevation be ‘θ’.

Then from the figure

∴ Angle of elevation of Sun rays with the ground = 60°.

You want to erect a pole of height 10 m with the support of three ropes. Each rope has to make an angle 30° with the pole. What should be the length of the rope?

Answer:

Height of the pole = 10 m

Let the length of each rope = x

Angle made by the rope with the pole = 30°

Suppose you are shooting an arrow from the top of a building at a height of 6 m to a target on the ground at an angle of depression of 60°. What is the distance between you and the object?

Answer:

Height of the building = 6 m

Angle of depression = Angle of elevation at the ground = 60°

Let the distance of the target from the shooting point = x m

Then from the figure

sin 60° =6/x

∴ Distance = 4√3 m or

4 × 1.732 = 6.928 m.

An electrician wants to repair an electric connection on a pole of height 9 m. He needs to reach 1.8 m below the top of the pole to do repair work. What should be the length of the ladder which he should use, when he climbs it at an angle of 60° with the ground? What will be the distance between foot of the ladder and foot of the pole?

Answer:

Height of the pole = 9m

Height of the point from the ground where he reaches the pole = 9 - 1.8 = 7.2 m

Angle of elevation = 60°

Angle of depression = Angle of elevation at the ground = 60°

Let the distance of the target from the shooting point = x m

Then from the figure

sin 60° =7.2/x

A boat has to cross a river. It crosses the river by making an angle of 60° with the bank of the river due to the stream of the river and travels a distance of 600 m to reach the another side of the river. What is the width of the river?

Answer:

Let the width of the river = AB = x m

Angle made by the boat = 60°

Distance travelled = AC = 600 m

From the figure

cos 60° =x/600

1/2 = x/600

⇒ x = 600/2 = 300 m

In the figure

A = Boat’s place

C = Reach place of another side (or) Point of observation.

AC = Travelling distance of the boat ∠AC = 60°

AB = width of the river AB

In △ABC, sin 60° =AB/AC

⇒ √3/2 = AB/600

⇒ AB = 600 × √3/2 = 300√3

An observer of height 1.8 m is 13.2 m away from a palm tree. The angle of elevation of the top of the tree from his eyes is 45°. What is the height of the palm tree?

Answer:

Height of the observer = 1.8 m

Distance of the observer from the palm tree = 13.2 m

From the figure

tan 45° = x/13.2

⇒ 1 = x/13.2

⇒ x = 13.2 m

∴ Height of the palm tree = 13.2 + 1.8 = 15 m.

In the given figure, AC = 6 cm, AB = 5 cm and ∠BAC = 30°. Find the area of the triangle.

Answer

Draw a perpendicular BD to AC

∴ BD ⊥ AC

Now let AD = 6 - x and DC = x

Given AB = 5 cm and ∠BAD = 30° then in △ABD

sin 30° = BD/AB = BD/5 = 1/2

⇒ BD = 52 = 2.5 cm

= 12 × AC × BD

= 12 × 6 × 2.5

= 7.5 cm^{2}

A TV tower stands vertically on the side of a road. From a point on the other side directly opposite to the tower, the angle of elevation of the top of tower is 60°. From another point 10 m away from this point, on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the road.

Answer:

Let the height of the tower = h mts say

Width of the road be = x m.

Distance between two points of observation = 10 cm.

Angles of elevation from the two points = 60° and 30°.

From the figure

tan 60° =h/x

√3 × √3x = 10 + x

⇒ 3x - x = 10

⇒ 2x = 10

⇒ x = 10/2 = 5m

∴ Width of the road = 5 m

Now Height of the tower = √3x = 5√3 m.

A 1.5 m tall boy is looking at the top of a temple which is 30 metre in height from a point at certain distance. The angle of elevation from his eye to the top of the crown of the temple increases from 30° to 60° as he walks towards the temple. Find the distance he walked towards the temple.

Answer:

Height of the temple = 30 m

Height of the man = 1.5 m

Initial distance between the man and temple = d m. say

Let the distance walked = x m.

From the figure

tan 30° = 30-1.5/d

= 19 × 1.732

= 32.908 m.

∴ Distance walked = 32.908 m.

A statue stands on the top of a 2 m tall pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the statue.

Answer:

Height of the pedestal = 2 m.

Let the height of the statue = h m. Angle of elevation of top of the statue = 60°.

Angle of elevation of top of the pedestal = 45°.

Let the distance between the point of observation and foot of the pedestal = x m.

From the figure

tan 45° =2/x

1 =2/x

∴ x = 2 m.

Also tan 60° =2+h/x

⇒ √3 =2+h/x

⇒ 2√3 = 2 + h

⇒ h = 2√3 - 2

= 2(√3-1)

= 2(1.732 - 1)

= 2 × 0.732

= 1.464 m.

From the top of a building, the angle of elevation of the top of a cell tower is 60° and the angle of depression to its foot is 45°. If distance of the building from the tower is 7 m, then find the height of the tower.

Answer:

Angle of elevation of the top of the tower = 60°.

Angle of depression to the foot of the tower = 45°.

Distance between tower and building = 7 m.

Let the height of the building = x m and tower = y m.

From the figure

tan 45° =x/7

1 =x/7

∴ x = 7 m.

Also tan 60° =y?x/7

⇒ √3 =y?x/7

⇒ 7√3 = y ? 7

∴ y = 7 + 7√3

= 7 (√3 + 1)

= 7(1.732 + 1)

= 2.732 × 7

= 19.124 m.

A wire of length 18 m had been tied with electric pole at an angle of eleva¬tion 30° with the ground. As it is covering a long distance, it was cut and tied at an angle of elevation 60° with the ground. How much length of the wire was cut ?

Answer:

Length of the wire = 18 m

Let the length of the wire removed = x

Height of the pole be = h

From the figure

sin 30° =h/18

⇒ 1/2=h/18

⇒ h =18/2= 9 m

Also sin 60° =h/18?x

= 6√3(√3-1)

= 6(3-√3)

= 18 - 6√3

= 18 - 10.392

= 7.608 m.

The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 30 m high, find the height of the building.

Answer:

Height of the tower = 30 m

Angle of elevation of the top of the tower = 60°.

Angle of elevation of the top of the building = 30°.

Let the distance between the foot of the tower and foot of the building be d m and height of the building be x m.

From the figure

tan 60° =30/d

√3 =30/d

Two poles of equal heights are standing opposite to each other on either side of the road, which is 120 feet wide. From a point between them on the road, the angles of elevation of the top of the poles are 60°and 30° respectively. Find the height of the poles and the distances of the point from the poles.

Answer:

Width of the road = 120 f.

Angle of elevation of the top of the 1st tower = 60°.

Angle of elevation of the top of the 2 tower = 30°.

Let the distance of the point from the 1st pole = x.

Then the distance of the point from

the 2nd pole = 120 - x.

and height of each pole = h say.

From the figure

tan 60° =h/x

⇒ √3 =h/x

⇒ h = √3x ……..(1)

Also tan 30° =h/120?x

⇒ √3.√3x = 120-x

⇒ 3x = 120 - x

⇒ 3x + x = 120

⇒ 4x = 120

⇒ x = 120/4 = 30 ft

Now h = √3x = √3 × 30 = 1.732 x 30 = 51.960 feet

∴ Distances of the poles = 30 ft. and 120 - 30 fts = 90 ft.

Height of each pole = 51.96 ft.

The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m, find the height of the tower from the base of the tower and in the same straight line with it are complementary.

Answer:

Let the height of the tower = h m.

Angles of elevation of the top of the tower from two points = x° and (90° - x)

From the figure

tan x =h/4……. (1)

Also tan (90° - x) =h/9

⇒ cot x =h/9

⇒1/tanx = h/9

∴ tan x = 9/h …….. (2)

From (1) and (2)

tan x =h/4=9/h

∴h/4=9/h

h × h = 9 × 4

⇒ h^{2}= 36

⇒ h = 6 m

The angle of elevation of a jet plane from a point A on the ground is 60°. After 4 flight of 15 seconds, the angle of elevation changes to 30°. If the jet plane is flying at a constant height of 1500√3 meter, find the speed of the jet plane. (√3 = 1.732)

Answer:

Height of the plane from the ground PM = RN = 1500√3 m.

Angle of elevation are 30° and 60°.

From the figure

tan 60° = PM/QM

QM + MN = 1500√3 × √3

1500 + MN = 1500 × 3

MN = 4500 - 1500

MN = 3000 mts.

∴ Distance travelled in 15 seconds = 3000 mts.

∴ Speed of the jet plane = distance/time =3000/15 = 200 m/s

= 200 × 18/5 kmph

= 720 kmph

Speed = 200 m/sec. or 720 kmph.

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