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Maths project some applications of trignometry- class10 ppt

ITS ALL ABOUT APPLICATION OF trignometry.....................

ITS ALL ABOUT APPLICATION OF trignometry.....................

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Maths project some applications of trignometry- class10 ppt

1. 1. MADE BY :- ULTRON SOME APPLICATION OF TRIGNOMETRY
2. 2. INTRODUCTION • TRIGONOMETRY IS THE BRANCH OF MATHEMATICS THAT DEALS WITH TRIANGLES PARTICULARLY RIGHT TRIANGLES. FOR ONE THING TRIGONOMETRY WORKS WITH ALL ANGLES AND NOT JUST TRIANGLES. THEY ARE BEHIND HOW SOUND AND LIGHT MOVE AND ARE ALSO INVOLVED IN OUR PERCEPTIONS OF BEAUTY AND OTHER FACETS ON HOW OUR MIND WORKS. SO TRIGONOMETRY TURNS OUT TO BE THE FUNDAMENTAL TO PRETTY MUCH EVERYTHING
3. 3. BASIC FUNDAMENTALS •
4. 4. Angle of Depression:
5. 5. 600 300 600 a 2a2a 1 1 1 45 2 1 45 2 1 45    Tan Cos Sine
6. 6. B A C Sin / Cosec  P (pandit) H (har) Cos / Sec  B (badri) H (har) Tan / Cot  P (prasad) B (bole) This is prett y easy! BASE (B) PERPENDICULAR (P)  7
7. 7.  A 0 30 45 60 90 Sin A 0 1 Cos A 1 0 Tan A 0 1 Not Defined Cosec A Not Defined 2 1 Sec A 1 2 Not Defined Cot A Not Defined 1 0 8
8. 8. The angle of elevation of the top of a tower from a point At the foot of the tower is 300 . And after advancing 150mtrs Towards the foot of the tower, the angle of elevation becomes 600 .Find the height of the tower 150 h d 30 60 mh h hh hh hh dofvaluethengSubstituti hd From hdFrom d h Tan d h Tan 9.129732.1*75 31502 31503 31503 )1503(3 .......... )150(3 )2( 3)1( )2( 150 360 )1( 3 1 30           
9. 9. ? 45o ? What you’re going to do next? Heights and Distances
10. 10. In this situation , the distance or the heights can be founded by using mathematical techniques, which comes under a branch of ‘trigonometry’. The word ‘ trigonometry’ is derived from the Greek word ‘tri’ meaning three , ‘gon’ meaning sides and ‘metron’ meaning measures. Trigonometry is concerned with the relationship between the angles and sides of triangles. An understanding of these relationships enables unknown angles and sides to be calculated without recourse to direct measurement. Applications include finding heights/distances of objects.
11. 11. Sun’s rays casting shadows mid-afternoon Sun’s rays casting shadows late afternoon An early application of trigonometry was made by Thales on a visit to Egypt. He was surprised that no one could tell him the height of the 2000 year old Cheops pyramid. He used his knowledge of the relationship between the heights of objects and the length of their shadows to calculate the height for them. (This will later become the Tangent ratio.) Can you see what this relationship is, based on the drawings below? Thales of Miletus 640 – 546 B.C. The first Greek Mathematician. He predicted the Solar Eclipse of 585 BC. Trigonometry Similar Triangles Similar Triangles Thales may not have used similar triangles directly to solve the problem but he knew that the ratio of the vertical to horizontal sides of each triangle was constant and unchanging for different heights of the sun. Can you use the measurements shown above to find the height of Cheops? 6 ft 9 ft 720 ft h 6 720 9 h  480 ft (Egyptian feet of course)4 6 720 9 80 ft x h  
12. 12. h Early Applications of Trigonometry Finding the height of a mountain/hill. Finding the distance to the moon. Constructing sundials to estimate the time from the sun’s shadow.
13. 13. Historically trigonometry was developed for work in Astronomy and Geography. Today it is used extensively in mathematics and many other areas of the sciences. •Surveying •Navigation •Physics •Engineering
14. 14. 45o Angle of elevation A C B In this figure, the line AC drawn from the eye of the student to the top of the tower is called the line of sight. The person is looking at the top of the tower. The angle BAC, so formed by line of sight with horizontal is called angle of elevation. Tower Horizontal level
15. 15. 45o Mountain Angle of depression A B C Object Horizontal level
16. 16. 45o Angle of elevation A C B Tower Horizontal level Method of finding the heights or the distances Let us refer to figure of tower again. If you want to find the height of the tower i.e. BC without actually measuring it, what information do you need ?
17. 17. We would need to know the following: i. The distance AB which is the distance between tower and the person . ii. The angle of elevation angle BAC . Assuming that the above two conditions are given then how can we determine the height of the height of the tower ? In ∆ABC, the side BC is the opposite side in relation to the known angle A. Now, which of the trigonometric ratios can we use ? Which one of them has the two values that we have and the one we need to determine ? Our search narrows down to using either tan A or cot A, as these ratios involve AB and BC. Therefore, tan A = BC/AB or cot A = AB/BC, which on solving would give us BC i.e., the height of the tower.
18. 18. The angle of elevation of the top of a tower from a point At the foot of the tower is 300 . And after advancing 150mtrs Towards the foot of the tower, the angle of elevation becomes 600 .Find the height of the tower 150 h d 30 60 mh h hh hh hh dofvaluethengSubstituti hd From hdFrom d h Tan d h Tan 9.129732.1*75 31502 31503 31503 )1503(3 .......... )150(3 )2( 3)1( )2( 150 360 )1( 3 1 30            Example 1:-
19. 19. 45 BA CDE 60 I see a bird flying at a constant speed of 1.7568 kmph in the sky. The angle of elevation is 600. After ½ a minute, I see the bird again and the angle of elevation is 450. The perpendicular distance of the bird from me, now will be(horizontal distance) ? ANSWER : Let A be the initial position and B be the final position of the bird, <AED= 600 , <BED = 450 Let E be my position. Time required to cover distance from A to B=30 sec. Speed of bird= 1.7568 × m/s Distance travelled by bird in 30 sec. = 1.7568 × × 30 = 14.64 m In right angled = Tan 600 . Thus, ED = In right angled As EC=ED+DC ,,, BC= +DC ,,, BC= + 14.64 18 5 18 5 ED AD AED, 3 AD ECBCBCE  ,  3 AD  3 BC  64.14 3 1 1       BC               3 1 1 1 64.14BC 320 1732.1 3 64.14           m Example 2:-
20. 20. EXAMPLE 3:- 30 ° 30 ° Step 1: Let ‘x’ be the distance the airplane must fly to be directly above the tree. Step 2: The level ground and the horizontal are parallel, so the alternate interior angles are equal in measure. Step 3: In triangle ABC, tan 30=AB/x. Step 4: x = 2 / tan 30 Step 5: x = (2*31/2) Step 6: x = 3.464 So, the airplane must fly about 3.464 miles to be directly above the tree. D
21. 21. 24 I get it!
22. 22. Thank you X-B