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Start studying Applications of Trigonometry. Learn vocabulary, terms, and more with flashcards, games, and other study tools. TrigonometryAnalytic Trigonometry with Applications, 11th EditionCK-12 Trigonometry - Second EditionTrigonometryAnswers to Plane Trigonometry with Practical ApplicationsThe Elements of Plane Trigonometry, and Their Application to the Measurement of Heights and Distances, Etc. (A Table of Logarithms of Numbers from 1 to 10000.-A Traverse Table.

Tenth class mathematics chapter 12 exercises 12.1 and 12.2 solutions are given.

These solutions are very easy to understand. First study the chapter 12 well for many times.

Observe the example probems and solutions well. Try them in your own method. Next solve the exercise problems.

Observe the solutions and try them in your own method.

You can see the solutions for

1. Real numbers

2. Sets

3. Polynomials

4. Pair of linear equations in two variables

5. Quadratic equations

6. Progressions

7. Coordinate geometry

8. Similar triangles

9. Tangents and Secants to a Circle

10. Mensuration

Applications Of Trigonometry Class 10 Answers

11. Trigonometry

12. Applications of trigonometry

13. Probability

14. Statistics

Exercise 12.1 class 10 maths chapter 12

SSC Maths Applications of Trigonometry solutions

Exercise 12.2

Chapter 12 applications of trigonometry solutions class 10

Optional exercise

F

Notes : Observe the solutions and try them in your own methods.

You can see solutions for Inter Maths IIB

1. Circle

2. System of Circles

3. Parabola

4. Ellipse

5. Hyperbola

You can also see solutions for Inter Maths IIA

1. Complex numbets

2. De Moivre’s Theorem

3. Quadratic Equations 1

4. Theory of Equations

5. Permutations and Combinations

6. Binomial Theorem

7. Partial Fractions

8. Measures of Dispersion

9. Probability

10. Random variables and Probability Distribution

For examination purpose you can see

Question 1 :

The angle of elevation of the top of the building at a distance of 50 m from its foot on a horizontal plane is found to be 60 degree. Find the height of the building.

Question 2 :

A ladder placed against a wall such that it reaches the top of the wall of height 6 m and the ladder is inclined at an angle of 60 degree. Find how far the ladder is from the foot of the wall.

Question 3 :

A string of a kite is 100 meters long and the inclination of the string with the ground is 60°. Find the height of the kite, assuming that there is no slack in the string.

Question 4 :

From the top of the tower 30 m height a man is observing the base of a tree at an angle of depression measuring 30 degree. Find the distance between the tree and the tower.

Question 5 :

A man wants to determine the height of a light house. He measured the angle at A and found that tan A = 3/4. What is the height of the light house if A is 40 m from the base ?

Question 6 :

A ladder is leaning against a vertical wall makes an angle of 20° with the ground. The foot of the ladder is 3 m from the wall. Find the length of ladder.

Question 7 :

A kite is flying at a height of 65 m attached to a string. If the inclination of the string with the ground is 31°, find the length of string.

Question 8 :

The length of a string between a kite and a point on the ground is 90 m. If the string is making an angle θ with the level ground such that tan θ = 15/8, how high will the kite be ?

Question 9 :

An airplane is observed to be approaching the air point. It is at a distance of 12 km from the point of observation and makes an angle of elevation of 50 degree. Find the height above the ground.

Question 10 :

A balloon is connected to a meteorological station by a cable of length 200 m inclined at 60 degree angle with the ground. Find the height of the balloon from the ground. (Imagine that there is no slack in the cable)

Answers

Question 1 :

The angle of elevation of the top of the building at a distance of 50 m from its foot on a horizontal plane is found to be 60°. Find the height of the building.

Solution :

Now we need to find the length of the side AB.

tanθ = Opposite side/Adjacent side

tan 60° = AB/BC

√3 = AB/50

√3 x 50 = AB

AB = 50√3

Approximate value of √3 is 1.732

AB = 50 (1.732)

AB = 86.6 m

So, the height of the building is 86.6 m.

Question 2 :

A ladder placed against a wall such that it reaches the top of the wall of height 6 m and the ladder is inclined at an angle of 60°. Find how far the ladder is from the foot of the wall.

Solution :

Here AB represents height of the wall, BC represents the distance between the wall and the foot of the ladder and AC represents the length of the ladder.

In the right triangle ABC, the side which is opposite to angle 60° is known as opposite side (AB), the side which is opposite to 90° is called hypotenuse side (AC) and remaining side is called adjacent side (BC).

Now, we need to find the distance between foot of the ladder and the wall. That is, we have to find the length of BC.

tan θ = Opposite side/Adjacent side

tan60° = AB/BC

√3 = 6/BC

BC = 6/√3

BC = (6/√3) x (√3/√3)

BC = (6√3)/3

BC = 2√3

Approximate value of √3 is 1.732

BC = 2 (1.732)

BC = 3.464 m

So, the distance between foot of the ladder and the wall is 3.464 m.

Question 3 :

A string of a kite is 100 meters long and the inclination of the string with the ground is 60°. Find the height of the kite, assuming that there is no slack in the string.

Solution :

Now we need to find the height of the side AB.

Sin θ = Opposite side/Hypotenuse side

sinθ = AB/AC

sin 60° = AB/100

√3/2 = AB/100

(√3/2) x 100 = AB

AB = 50 √3 m

So, the height of kite from the ground 50 √3 m.

Question 4 :

From the top of the tower 30 m height a man is observing the base of a tree at an angle of depression measuring 30 degree. Find the distance between the tree and the tower.

Solution :

Here AB represents height of the tower, BC represents the distance between foot of the tower and the foot of the tree.

Now we need to find the distance between foot of the tower and the foot of the tree (BC).

tan θ = Opposite side/Adjacent side

tan 30° = AB/BC

1/√3 = 30/BC

BC = 30√3

Approximate value of √3 is 1.732

BC = 30 (1.732)

BC = 81.96 m

So, the distance between the tree and the tower is 51.96 m.

Question 5 :

A man wants to determine the height of a light house. He measured the angle at A and found that tan A = 3/4. What is the height of the light house if A is 40 m from the base ?

Solution :

Now we need to find the height of the light house (BC).

tanA = Opposite side/Adjacent side

tanA = BC/AB

Given : tanA = 3/4

3/4 = BC/40

3 x 40 = BC x 4

BC = (3 x 40)/4

BC = (3 x 10)

BC = 30 m

So, the height of the light house is 30 m.

Question 6 :

A ladder is leaning against a vertical wall makes an angle of 20° with the ground. The foot of the ladder is 3 m from the wall. Find the length of ladder.

Solution :

Now we need to find the length of the ladder (AC).

Cos θ = Adjacent side/Hypotenuse side

Cos θ = BC/AC

Cos 20° = 3/AC

0.9397 = 3/AC

AC = 3/0.9397

AC = 3.192

So, the length of the ladder is about 3.193 m.

Question 7 :

A kite is flying at a height of 65 m attached to a string. If the inclination of the string with the ground is 31°, find the length of string.

Solution :

Now we need to find the length of the string AC.

Sin θ = Opposite side/Hypotenuse side

Sin θ = AB/AC

Sin 31° = AB/AC

0.5150 = 65/AC

AC = 65/0.5150

AC = 126.2 m

Hence, the length of the string is 126.2 m.

Question 8 :

The length of a string between a kite and a point on the ground is 90 m. If the string makes an angle θ with the ground level such that tan θ = 15/8, how high will the kite be ?

Solution :

Now we need to find the length of the side AB.

Tan θ = 15/8 --------> cot θ = 8/15

csc θ = √(1+ cot²θ)

csc θ = √(1 + 64/225)

csc θ = √(225 + 64)/225

csc θ = √289/225

csc θ = 17/15 -------> sin θ = 15/17

But, sin θ = Opposite side/Hypotenuse side = AB/AC

AB/AC = 15/17

AB/90 = 15/17

AB = (15 x 90)/17

AB = 79.41

So, the height of the tower is 79.41 m.

Question 9 :

An airplane is observed to be approaching a point that is at a distance of 12 km from the point of observation and makes an angle of elevation of 50 degree. Find the height of the airplane above the ground.

Solution :

Now we need to find the length of the side AB.

From the figure given above, AB stands for the height of the airplane above the ground.

sin θ = Opposite side/Hypotenuse side

sin 50° = AB/AC

0.7660 = h/12

0.7660 x 12 = h

h = 9.192 km

So, the height of the airplane above the ground is 9.192 km.

Question 10 :

A balloon is connected to a meteorological station by a cable of length 200 m inclined at 60 degree angle with the ground. Find the height of the balloon from the ground. (Imagine that there is no slack in the cable)

Solution :

Now we need to find the length of the side AB.

From the figure given above, AB stands for the height of the balloon above the ground.

sin θ = Opposite side/Hypotenuse side

sin θ = AB/AC

sin 60° = AB/200

√3/2 = AB/200

AB = (√3/2) x 200

AB = 100√3

Approximate value of √3 is 1.732

AB = 100 (1.732)

AB = 173.2 m

Applications Of Right Triangle Trigonometry Answers

So, the height of the balloon from the ground is 173.2 m.

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