# Applications Of Trigonometry

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## Presentation Description

Made By- Munish Sharma TGT-Maths J.N.V. Una Himachal Pradesh

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## Presentation Transcript

### PowerPoint Presentation:

Applications of trigonometry Class : X Prepared by : Munish Sharma TGT-Maths Name Of topic

### PowerPoint Presentation:

In this topic we shall make use of trigonometric ratios to find the height of a tree, a tower, a water tank, width of a river, distance of a ship from light house etc.

### Line Of Sight:

Line Of Sight We observe generally that children usually look up to see an aeroplane when it passes overhead. The line joining their eyes to the plane, while looking up is called the line of sight.

### Line Of Sight:

Line Of Sight Line of sight

### Angle of elevation:

Angle of elevation The angle which the line of sight makes with a horizontal line drawn away from their eyes is called the angle of elevation of the aeroplane from them.

### PowerPoint Presentation:

Line of sight Angle of elevation Angle of elevation Horizontal

### Angle of depression:

Angle of depression If the pilot of the aeroplane looks downwards at any object on the ground then the angle between his line of sight and horizontal line drawn away from his eyes is called the angle of depression.

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Line of sight Angle of depression Angle of depression Horizontal Object

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Lets discuss some problems related to the topic from our daily life Applications of Heights & Distances

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Example 1. A tower stands on the ground. The angle of elevation from a point on the ground which is 30 metres away from the foot Of the tower is 30⁰. Find the height of the tower. (Take √3 = 1.732) 30⁰ 30 m h A B C Let AB be the tower h metre high. Let C be a point on the ground which is 30 m away from point B, the foot of the tower. So BC = 30 m Then ACB = 30⁰ Now we have to find AB i.e. height ‘h’ of the tower . Solution .

### In right angled triangle ABC:

In right angled triangle ABC Sin θ Cos θ Tan θ Pandit Badri Prasad Hari Hari Bole Cosec θ Sec θ Cot θ AB BC 1 √3 h 30 h 30 √3 30 x √3 √3 x √3 30 x √3 3 10√3 m 10 x 1.732 m 17.32 m Hence, height of the tower = 17.32 m tan 30⁰ 30⁰ 30 m h A B C Now we shall find the trigonometric ratio combining AB and BC .

### PowerPoint Presentation:

Example 2. A tree stands vertically on the bank of a river. From a point on the other end directly opposite the tree, the angle of elevation of the top of the tree is 60⁰, from a point 20 metres behind this point on the same bank ,the angle of elevation of the top of the tree is 30 ⁰. Find the height of the tree and width of the river. (Take √3 = 1.732) Let PQ = h metres be the height of the tree. Let A be a point on the ground and B is another point on the ground which is 20m behind point A. So BA = 20 mtr Then PAQ = 60⁰ Let AQ = x mtr be the width of the river 30⁰ 20 m P B A Q x 60⁰ h Solution . And PBQ = 30⁰

### In right angled triangle AQP:

In right angled triangle AQP Sin θ Cos θ Tan θ Pandit Badri Prasad Hari Hari Bole Cosec θ Sec θ Cot θ PQ AQ tan 60⁰ √3 1 h x h √ 3 x √ 3 X √ 3 x Hence, width of the river = 10 m The trigonometric ratio combining PQ and AQ is tan Ө In right angled triangle PQB tan 60⁰ PQ BQ tan 30⁰ 1 √3 h x + 20 x + 20 √ 3 h 3x x + 20 x + 20 2x 20 x 10 30⁰ 20 m P B A Q x 60⁰ h

### PowerPoint Presentation:

√ 3 x h Put x 10 √ 3 X 10 h 1.732 X 10 h 17.32 mtr h Hence, height of the tree = 17.32 m

### In right angled triangle PBQ:

Example 3. As observed from the top of a light house , 100m above Sea level, the angle of depression of a ship, sailing towards it 30 ⁰ to 45 ⁰. Determine the distance travelled by the ship during the period of observation. (Take √3 = 1.732) 30⁰ X P B A Q y 45⁰ 100 m 45⁰ 30⁰ X Horizontal Top of light house P is the top of the light house PQ. Solution . PQ = 100 m PX is horizontal line through P And PX || AQ. So APX = PAQ = 30⁰ and BPX = PBQ = 45⁰ In right angled triangle PBQ PQ BQ Tan 45⁰

### In right angled triangle PBQ:

100 y 1 √3 1 y = 100 In right angled triangle PBQ PQ BQ Tan 30⁰ 100 x+ y x + y = 100 √3 x + y = 100 X 1.732 x + 100 = 173.2 Because y = 100 X = 173.2 -100 X = 73.2 Hence, distance travelled by the ship = 73.2 m

### PowerPoint Presentation:

An electric pole is 12 m high. A steel wire is tied from the top of the pole to a point on the ground to keep it upright. If the wire makes an angle of 60⁰ with the horizontal . Then find the height of the pole (Take √3 = 1.732) Home Work 2. Two pillars of equal height are on either side of a road, which is 100m . At a point on the road between the pillars, the angle of elevation of the top of the pillars are 60⁰ and 30⁰ respectively. Find the position of the point between the pillars and height of each pillar. 3. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60⁰ and the angle of depression of its foot is 45⁰. Find the height of the tower .(Take √3 = 1.732)

### END OF SHOW:

END OF SHOW Presented By : Munish Sharma 