Selection File type icon File name Description Size Revision Time User; ĉ: Chapter 3 Answer Keys.docx View Download: 89k: v. 2: Nov 5, 2012, 1:09 PM: nesterling. Class 10 Chapter 9 some application of trigonometry is an important topic to discuss as it tells how trigonometry is used to find the height and distance of different objects such as the height of the building, the distance between the Earth and Planet and Stars, the height of the highest mountain Mount Everest, etc. The following videos shows more examples of solving application of trigonometry word problems. Example 1: Suppose that a 10 meter ladder is leaning against a building such that the angle of elevation from ground to the building is 62 degrees. Find the distance of the foot of the ladder from the wall. Revise NCERT Solutions for CBSE Class 10 Mathematics Chapter 9 Some Applications of Trigonometry to hone your problem-solving skills to ace the board exam. When you practise the textbook solutions, you relearn the methods for applying trigonometry concepts. Solve problems wherein you must calculate the distance between two objects such as ships.

Trigonometry: Trigonometric Applications. Correct answer: Explanation: Since the triangle in question is a right triangle we can use the Pythagorean Theorem. Refer to the lesson on how to solve trigonometry applications for more information and examples. Fill in all the gaps, then press 'Check' to check your answers. Use the 'Hint' button to get a free letter if an answer is giving you trouble. You can also click on the ' ?' button to get a clue.

Question 1 :

Application-trigonometry-answers 2/6 Downloaded from www.liceolefilandiere.it on December 14, 2020 by guest Applications - Radford SOME A PPLICATIONS OF TRIGONOMETRY Trigonometry Applications of trigonometry - ibiblio Application Trigonometry Answers 11.1 Applications of Sinusoids Applications of Trigonometry Chapter.

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 it makes an angle of 60° with horizontal. 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 :

He 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 flying at a height of 65 m is attached to a string inclined at 31° to the horizontal. What is 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 . 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 degree. 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 degree. 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 degree is known as opposite side (AB), the side which is opposite to 90 degree 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 it makes an angle of 60° with horizontal. 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

Applications Of Right Triangle Trigonometry Answers

BC = (3 x 10)

BC = 30 m

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

Question 6 :

A man wants to determine the height of a light house. He 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

0.9396 = 3/AC Antares autotune 5 crack.

AC = 3/0.9396

AC = 3.192

So, the length of the ladder is 3.192 m.

Question 7 :

A kite flying at a height of 65 m is attached to a string inclined at 31° to the horizontal. What is 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 is making an angle θ with the level ground 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 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.

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 . 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

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

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