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12.3 Example : A 1.5 m tall girl is looking the top of the tree from a point at certain distance. The angle of elevation from her eye to the top of the tree 30 0 . The angle of elevation changes to 45 0 when a girl move towards the tree. Find the height of the tree.

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  1. 12.3 Example : A 1.5 m tall girl is looking the top of the tree from a point at certain distance. The angle of elevation from her eye to the top of the tree 300 . The angle of elevation changes to 450when a girl move towards the tree. Find the height of the tree. AB = AE + EB = height of the tree CD = height of the girl =1.5m ADE = The angle of elevation observed from the point D to the top of the tree = 300 CP =DQ = Distance moved by girl towards tree =11 m AQE= The angle of elevation observed after moving 11 m towards the tree to top of the tree = 450 CD = PQ = BE =1.5m QE =PB = Distance between observing point Q to foot of the tree = Let x m AE = Let h meters 11m In ΔAQE, Opposite side to Q is AE = h , Adjacent side to Q is QE = x0 1.5m 1.5m 11m

  2. In ΔADE, Opposite side to D is AE=h , Adjacent side to D is DE = 11 + x AB = AE + EB = AB = h+1.5 =15.026+1.5 = 16.526m height of the tree = 16.526 m

  3. Example – 6 Two men on either side of a temple of 30 meters height observe its top at the angles of elevation 300 and 600respectively. Find the distance between the two men. AB = height of the temple = 30 m EG = DF = distance between two men AEB = The angle of elevation observed the top of the temple by first man = 300 30 m AGB = The angle of elevation observed the top of the temple by second man =600 EB = Distance between first man and temple = Let x meters GB = The Distance between second man and temple = Let y meters

  4. In ΔAEB, Opposite side to E is AB, Adjacent side to E is EB In ΔAGB, Opposite side to G is AB , Adjacent side to G is BG EG= x + y = Distance between two men m

  5. Example – 7 A Straight highway leads to the foot of a tower. Ramaiah standing at the top of the tower observes a car at angle of depression 300 . The car is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 600. Find the time taken by the car to reach the foot of the tower from this point . AB = height of the tower = Let h meters EBC = BCA = The angle of depression Ramaiah standing on the top of the tower observes a car = 300 CD =Distance travelled by car in 6 Seconds = Let x meters The angle of depression observed Six seconds later to the uniform speed car =600 Distance to be travelled to reach the foot of the tower = Let d meters

  6. In ΔABD, Opposite side to D is AB =h , Adjacent side to D is AD = d Time taken to travel x meter = 6 Seconds Time taken to travel x / 2 meters = 6/2 = 3 Seconds In ΔABC, Opposite to C is AB = h , Adjacent side to C is AC = x +d The time taken by the car to reach the foot of the tower from the point = 3 Seconds

  7. Exercise -12.2 1.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 the tower is 600.Fromanother 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 300 . Find the height of the tower and the width of the road . AB = height of the tower AC = Width of the road ACB = The angle elevation of the top of the tower , observed from a point on the other side of the road =600 CD = The distance between the observing point C and Oberving point D which is on the line joining the foot of the tower = 10 m ADB =The angle of elevation from the Observing point D which is on the line joining the foot of the tower = 300 In ΔABC, Opposite side of C is AB=h , Adjacent side of C is AC=x 600 300 10 Ò$.

  8. In ΔABD, Opposide to D is AB=h, Adjacent side to D is AD is AD = x+10 AC = Width of the road =x = 5 meters AB = height of the tower = h = 8.66 meters

  9. Exercise - 12.2 2. A 1.5 m. tall boy is looking at the top of a temple which is 30 meters 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 300 to 600as he walks towards the temple. Find the distance he walked towards the temple . CD = Height of the boy =1.5m AB = Height of the temple = 30 m BDE = The angle of elevation Observed by boy at C to the top of the the temple =300 CP = The distance from the observing point C to Observing point P = Let x meters BQE = The angle of elevation observed by boy after walking from C to P to the top of the temple =600 30 m 28.5 m Let EQ = AP = y, QD =PC= x and CD=PQ=AE=1.5 m 600 300 1.5 m 1.5m BE = AB – AE = 30 – 1.5 = 28.5 m

  10. In Δ BQE , Opposite side to Q is AB= 28.5 m and Adjacent side to Q isQE = y The distance , boy walked towards the temple = 32.908 m In Δ BDE , Opposite side to D is AB=28.5 m , Adjacent side D is DE = DQ+EQ = x + y

  11. Exercise - 12.2 3. A Statue stands on the top of a 2m tall pedestal. From a point on the ground , the angle of elevation of the top of the stature is 600and from the same point , the angle of elevation of the top of the pedestal is 450. Find the height of the statue . AB = Height of the pedestal = 2m BC = Height of the statue = Let h meters h m . AD = The distance between the observing point D on the ground to bottom of the pedestal A = Let x meters ADC =The angle of elevation from the observing point D to the top of the statue=600 ADB = The angle of elevation from the observing point D to the top of te pedestal = 450 2 m In Δ ABD , opposite side to D is AB=2 m , Adjacent side to D is AD = Let x 600 450

  12. In Δ ACD, Opposite side to D is AC = 2+h , Adjacent side to D is AD = x BC = Height of the statue = 1.464 m

  13. Exercise -12.2 4. From the top of a building , the angle of elevation of the top of the cell tower is 600 and the angle of depression to its foot is 450. If distance of the building from the tower is 7 m , then find the height of the tower . AB = Height of the building = Let x m CD = Height of the Cell tower = h meters AC = Distance between Building and cell tower = 7 m BE = Horizontal line = AC = 7m BDE = The angle of elevation from the top of a building to the top of the cell tower =600 7m EBC= BCA = The angle of depression from top of the building to the foot of the cell tower =450 450 7 m CE=AB =x

  14. In Δ ABC , Opposite side to C is AB = x , Adjacent side to C is AC = 7 m In Δ BED, Opposite side to B is DE = CD – CE = h – x , Adjacent side to B is BE = 7 m CD = Height of the cell tower = 19.124 m

  15. Exercise - 12.2 5. A Wire of length, 18 m had been tied with electrical pole at an angle of elevation 300with the ground. Because if was converting a long distance, it was cut and tied at an angle elevation of 600 with the ground. How much length of the wire was cut ? AB = height of the electrical pole = Let h meters BC = length of the wire = 18 m. BD = Length of the wire after cut CD = Distance between wire before and after cut AD = Distance between foor of the Electrical pole and wire after cut = Let y 18 Ò$. ACB= The angle of elevation of the wire before cut = 300 ADB= The angle of elevation of the wire after cut = 600

  16. In ΔABC, Opposite side to C is AB, Adjacent side to C is AC, hypotenuse is AC In ΔABD, Opposite side to D is AB , Adacent side to D isAD, hypotenuse is BD The length of the wire was cut = BC – BD = 18 – 10.392 = 7.608 meters

  17. Exercise – 12.2 6. The angle of elevation of the top of a building from the foot of the tower is 300 and the angle of elevation of the top of the tower from the foot of the building is 600 . If the tower is 30 m high. Find the height of the building . AB = height of the building = Let h meters CD = height of the cell tower = 30 m AC = Distance between the building and foot of the cell tower = Let x meters ACB = Angle of elevation observed from the foot of the cell tower to top of the building = 300 30 m. CAD = Angle of elevation observed from foot of the building to the top of the cell tower =600 300

  18. In ΔABC, Opposite side to C is AB = h , Adjacent side to C is AC = x hypotenuse is BC In ΔADC, Opposite side to A is CD = 30, Adjacent side to A is AC = x , hypotenuse is AD AB= height of the building = h = 10 meters

  19. Exercise -12.2 7. 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 angle of elevation of the top of the poles are 600 and 300 respectively. Find the height of the poles and the distance of the point from the poles. AB = CD = Two poles of equal heights =h m AC = Width of the Road = 120 m E is a point between two poles on the road AEB = Angle of elevation observed from E to B = 600 CED = Angle of elevation observed from E to D=300 CE = Distance between Observing point E to pole CD = Let x m 600 AE = Distance between Observing point E to Pole AB = AC – CE = 120 – x m 120 Ò$.

  20. In ΔCDE, Opposite side to E is CD = h , Adjacent side to E is EC = x , hypotenuse ED In ΔADE, Opposite to E is AB = h , Adjacent side to E is EC = 120 – x , hypotenuse to EB AB = CD = Two poles of equal heights = 51.96 m CE = Distance between Observing point E to pole CD = 90 m

  21. Exercise - 12.2 8. 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. AB = height of the tower = Let h meters C and D are the two observing points which are on the same straight lines . AC = Distance between foot of the tower and Observing point C = 4 m AD= Distance between foot of the tower and observing point D = 9 m ADB = Angle of elevation observed from observing point D to top of the tower = Let  h Ò$. ACD = Angle of elevation observed point C to top of the tower which is complement to =90 –  4 Ò$. 9 Ò$.

  22. In ΔABD, Opposite to D is AB = h, Adjacent side to D is AD = 9,hypotenuse is BD ......... Comparing and  ……….  In ΔABC, Opposite side to C is AB = h, Adjacent to side to AC = 4, hypotenuse is BC AB = height of the tower = 6m

  23. 9. The angle of elevation of a jet plane from a point A on the ground is 600. After a flight of 15 seconds, the angle of elevation changes to 300 . If the jet plane is flying at a constant height of Exercise - 12.2 meters, Find the speed of the jet plane . DAB= The angle of elevation of a jet plane observed from a Point A on the ground = 600 DAC = After flight of 15 seconds, the angle of elevation observed from point A on the ground = 300 BD =CE = Constant height of the jet plane flying from the ground = Ò$. DE = BC = Within 15 Seconds , distance travelled by jet plane = Let x AD = Distance between observing point A to Perpendicular line of jet plane D = Let y

  24. In ΔADB, Opposite side to A is BD, Adjacent side to A is AD, hypotenuse is AB Speed of the jet plane = Distance / Time Speed of the jet plane = 200 meters / Second In ΔAEC, Opposite side to A is CE, Adjacent side to A is AE, hypotenuse is AC DE = BC = Within 15 Seconds , distance travelled by jet plane = 3000 m

  25. Exercise &12.2 10. Chinky observes a tower PQ of height “h” from a point A on the ground. She moves a distance “d” towards the foot of the tower and finds that the angle of elevation is twice at A . After she moves 3d/4 towards foot of the tower and finals that the angle of elevation is 3 times at A . Prove that 36h2 =35d2 PQ = height of the tower = Let h meters PAQ = Angle of elevation from the observing point A to the top of the tower = Let  AB = Distance between the observing points A and B = Let d m PBQ = Angle of elevation from the observing point B after she moves d meters towards the foot of the tower = 2 BC = Distance between two observing points B and C = 3d / 4 m h Ò$. PCQ =Angle of elevation from the observing point C after she moves 3d/4 meters towards the foot of the tower = 3 d Ò$.

  26. Extrerior Angle = Sum of two interior angles In ΔBQC, According to Rule of Sin h Ò$. d Ò$. d Ò$.

  27. In ΔPBQ, Opposite side to B is PQ , Adjacent side to B is PB , hypotenuse BQ h Ò$. d Ò$. d Ò$.

  28. Optional Exercise 1. A 1.2 m tall girl spots a ballon moving with the wind in a horizontal line at a height of 88.2 m from the ground . The angle of elevation of the balloon from the eyes of the girl at any instant is 600 . After some time, the antle of elevation reduces to 300 . Find the distance travelled by the balloon during the interval . CE= Height of the balloon flying from the ground = 88.2 m AB = Height of the girl = 1.2m =ED 88.2Ò$. 88.2Ò$. AB = ED = HG 87 Ò$. AE = Distance between the girl and foot of the balloon E = Let x m =BD 1.2Ò$. 1.2Ò$. 1.2Ò$. CBD = Angle of elevation observed by the girls = 600 FH=CE= After some time , height of balloon flying from the ground = 88.2 m FBD = After some , Angle of elevation observed by the girls = 300 DG = After some time , distance travelled by balloon = Let y m = EH FG = FH – GH = 88.2 – 1.2 = 87m

  29. In ΔCBD, Opposite side to B is CD, Adjacent side to B is BD In ΔFBG, Opposite side to B is FG, Adjacent side to B is BG 88.2Ò$. 88.2Ò$. 87 Ò$. 1.2Ò$. 1.2Ò$. 1.2Ò$. DG = Distance travelled by balloon = EH =

  30. Optional Exercise 2.The angle of elevation of the top of a tower from the foot of the building is 300 and the angle of elevation of the top the building from the foot of the tower is 600 .What is the ratio of heights of tower and building. AB = Height of the building = x m CD = Height of the tower = h m AD =Distance between the foot of the building and foot of the tower = d m DAC = Angle of elevation of top of the tower from foot of the building = 300 BCA= Angle of elevation of top of the building from foot of the tower = 600 ó00 D m

  31. In ΔABC, Opposite side to C is AB, Adjacent side to C is AC In ΔADC, Opposite side to A is CD, Adjacent side to A is AC Ratio of heights of the Building and Tower ó00 Ratio of heights of the Building and Tower = 3:1 d Ò$.

  32. Optional Exercise 3.The angle of elevation of the top of a light house from 3 boats A,B and C in a straight line of same side of the light house are a , 2a and 3a respectively . If the distance between the boats A and B is m meters . Fing the height of the light house ? PQ = heigt of the light house = Let h m PAQ =Angle of elevation observed from boat A to the top of the light house = a AB= Distance between boats A and B= Let x m PBQ = Angle of elevation observed from boat B to the top of the light house = 2a BC = Distance between boats B and C = let y m h Ò$. PCQ = Angle of elevation observed from boat C to the top of the light house = 3a X m Y m

  33. Exterior Angle = Sum of the Interior Angles In ΔBQC, as per Sin rule h Ò$. x Ò$. x Ò$. y Ò$.

  34. In ΔPBQ, Opposite side to B is PQ, Adjacent side to B is PB, hypotenuse is BQ h Ò$. x Ò$. x Ò$. y Ò$.

  35. Optional Exercise 4.Inner part of a cupboard is in the cuboidical shape with its length , breadth and height in ratio What is the angle made by the longest stick which can be inserted cupboard with its base inside . Inner part of a cupboard is in the cuboid shape Inner length of cupboard be AB, height be BC and the longest stick which can be inserted inside be AC 1 unit unit Angle made by stick which is inserted inside cupboard with its base be  1 unit In ΔABC, Opposide to  is BC, Adjacent side to  AB, hypotenuse is AC Angle made by longest stick which can be inserted inside the cupboard with its base = 450

  36. Optional Exercise 5.An Iron sperical ball of volume 232848 cm3 has been melted and converted into cone with the vertical angle of 1200 . What are its height and base ? AB = height of the cone = h cm BC = radius of the base of the cone = r cm AC = slant height of the cone = 232848 cm3 Vertical Angle of the cone = 1200 Vertical Angle bisector of cone = AB BAC = 600 232848 cm3 Volume of an iron sperical ball = 232848cm3 = Volume of the cone = 232848 cm3 In ΔABC, Opposite side to 600 is BC, Adjacent side to AB, hypotenuse AC

  37. Volume of the cone = 232848 cm3 cm cm

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