Applications of Trigonometry An Imagination line joining the top most point of the object and observer eye is called “ Line of Sight ”. OR The “ Line of Sight ” is the line drawn from the eye of an observer to a point on the object being viewed by the observer . The “ Horizontal line ” is the Parellel line drawn from the eye of an observer to object being viewed by the observer . Line of Sight Horizontal line

Line of Sight Angle of Elevation Horizontal line The “ Angle of Elevation of an object viewed ” is the angle formed by the line of sight with horizontal when it is obove the horizontal line OR The line of Sight is obove the horizontal line and agle between the line of sight and the horizontal line is called “ Angle of Elevation ” Line of Sight Angle of Elevation Horizontal line

Horizontal Line Angle of Depression Line of Sight The “ Angle of Depression of an object viewed ” is the angle formed by the line of sight with horizontal when it is below the horizontal line Horizontal Line OR Angle of Depression The line of Sight is below the horizontal line and angle between the line of sight and horizontal line is called angle of Depression . Line of Sight

These Modern Machines Which are used in the process of Survey are Theodolites. These machines work on the basis of Trigonometric ratioes.

Historical Information Trigonometry has been used by Surveyors for centries. They use to measure angles of elevations and depressions in the process of Survey. In the 19th Century, two large Theodolites were built by British India for the surveying project “ great trigonometric Survey” . During the Survey in 1852, the highest mountain peak in the World was discovered in the Himalayas. From the distance 160 km, the peak was observed from six different stations and the height of the peak was calculated. In 1856 , this peak was named after Sir George Everest , Who had Commissioned and first used the giant Theodolites. Those Theodolites are kept in the museum of the Survey of India in Dehradun for display.

The top of a clock tower is observed at an angle of elevation α0 and the foot of the tower is at distance of d meters from the observer. Draw the diagram for this data. Example : -1

Rinky observers a flower on the ground from the balcony of the first floor of a building at an angle of depression β0. The height of the first floor of the building is x meters.Draw the diagram for this data . Example : - 2 Angle of Depression

Example: 3 A large balloon has been tied with a rope and it is floating in the air . A Person has observed the balloon from the top of a building at angle of elevation of 1 and foot of the rope at angle of depression of 2 . The height of the building is h feet. Draw the diagram for this data .

1. Draw diagram for the following situations Do this 1. Draw diagram for the following situations ( i ) A Person is flying a kite at an angle of elevation α and the length of thread from his hand to kite is

Do this ( ii ) A Person observes two banks of a river at angle of depression 1 and 2 ( 1< 2 ) from the top of a tree of height h which is at a side of the river. The width of the river is “ d ”.

Think - Discuss Height of the school building = AB = d.tan α 1. You are observing top of your school building at an angle of elevation α from a point which is at d meter distance from foot of the building. Which trigonometric ratio would you like to consider to find the height of the School building. Think - Discuss Height of the school building = AB = d.tan α To find the height of the school building , We consider the Tan trigonomeric Ratio.

Think - Discuss 2. A ladder of length x meter is leaning against a wall making angle  with the ground. Which trigonometric ratio would you like to consider to find the height of the point on the wall at which the ladder is touching . The height of the point on wall at which ladder is touching = = AB = x.Sin  Sin Trigonometric Ratio would be considered to find the height of the point on the wall at which ladder is touching .

Example : 4 A boy observed the top of an electric pole at an angle of elevation of 600 when the observation point is 8 meters away from the foot of the pole . Find the height of the pole. AB = Let the height of the Electrical Pole = h meters BC = The Distance between the Electrical pole and Boy = 8 meters BCA = The angle of Elevation observed the top of the Electrical pole by the boy = 600 In ΔABC, Opposite side to C is AB, Adjacent side to C is BC and hypotenuse is AC AB = height of the pole Meters 8 Ò$.

Rajender observes a person standing on the ground from a helicopter at an angle of depression 450. If the helicopter flies at a height of 50 meters from the ground. What is the distance of the person from Rajender ? Example: 5 AB = height of the helicopter flies from the ground = 50 meters AC = The distance between the person on the ground to Rajender in helicopter = Let x meters 50 meters PAC= ACB= The angle of depression from Rajender to the person on the ground = 450 In ΔABC, Opposite Side to C is AB, Adjacent side to C is BC and hypotenuse is AC The Distance of the person from Rajender

Exercise : 12.1 1.A tower stands vertically on the ground . From a point which is 15 meters away from the foot of the tower. The angle of elevation of the top of the top of the tower is 450. What is the height of the tower? AB = The height of tower stands vertically on the ground = Let h meters BC = The distance between the observing point to foot of the tower = 15 meters 15 Ò$. BCA = The angle of elevation from the observing point to the top of the tower = 450 In ΔABC, Opposite side to C is AB , Adjacent side to C is BC and hypotenuse is AC Meters

Exercise - 12.1 2. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground by making 300 angle with the ground. The distance between the foot of the tree and the top of the tree on the ground is 6 cm. Find the height of the tree before falling down. AB = The height of the tree before fallen down A tree breaks due to storm at C and broken part bends so that the top of the tree touches the ground at D by making 300. CB= CD AD= The distance between foot of the tree to the top of the tree on the ground after falling down= 6 meters ADC= The angle of elevation broken part bends so that the top of the tree touches the ground = 300 In ΔADC, Opposite side to D is AC , Adjacent side to D is AD and hypotenuse is CD. 6 meters

meters In ΔADC In ΔADC AB = height of the tree before falling down ( CB = CD ) meters 6 meters

Exercise - 12.1 3. A Contractor wants to set up a slide for the children to play in the park . He wants to set it up at the height of 2 m and by making an angle of 300 with the ground . What should be the length of the slide ? AB = height of the slide = 2 m AC = Length of the slide ACB =The angle of elevation from the ground to top of the slide = 300 2 m In ΔABC, Opposite side to C is AB , Adjacent side to C is BC and hypotenuse is AC AC = Length of the slide = 4 m

4. Length of the shadow of a 15 meters high pole is Exercise -12.1 4. Length of the shadow of a 15 meters high pole is meters at 7 o’ clock in the morning. Then, what is the angle of elevation of the sun rays with the ground at the time ? AB = Height of the pole = 15 m AC = Length of the Shadow of the pole = m. BCA = The angle of elevation of the sun rays with the ground = Let  15m In ΔABC, Opposite side to C is AB , Adjacent side to C is AC and hypotenuse is BC m BCA= The angle of elevation of the sun rays with the ground = 600

AB = height of the pole = 10 m AC = length of the rope Exercise - 12.1 5. You want to erect a pole of height 10 m with the support of three ropes . Each rope has to make an angle 300 with the rope . What should be the length of the rope ? AB = height of the pole = 10 m AC = length of the rope ACB= The angle of elevation made by rope = 300 In ΔABC, Opposite side to C is AB, Adjacent side to C is BC and hypotenuse is AC 10 Ò$. AC = The length of the rope = 20 m

Exercise - 12.1 6. Suppose you are shooting an arrow from the top of a building at an height of 6 cm to a target on the ground at an angle of depression of 600 . What is the distance between you and the object ? AB = Height of the Building from the ground = 6 m XAC= ACB = The angle of depression from top of the building to a target = 600 6 Ò$. AC = Distance between object to me = ΔABC ÌZ C Äñý$$MæüP G§æþ$sìý ¿æý$f… AB ,Bçܯèþ² ¿æý$f… BC, MæüÆæÿ~… AC

DCB=The angle of elevation when use ladder with ground = 600 Exercise - 12.1 7. An electrician wants to repair an electric connection on a pole of height 9 m . He needs to reach 1.8 m below the top of the pole to do repair work. What should be the length of the ladder which he should use. When he climbs it at angle of 600 with the ground? What will be the distance between foot of the ladder and foot of the pole ? AB = Height of the Electric pole = 9 m AD = Electrician needs to reach below the top of the pole to do repair work =1.8 m BD = height of the pole to repair work = 9 - 1.8 = 7.2 m 1.8 Ò$. CD = length of the ladder to repair work BC = The distance between foot of the ladder and foot of the ladder 9 Ò$. DCB=The angle of elevation when use ladder with ground = 600 7.2 Ò$. In ΔBDC , Opposite side to C is BD, Adjacent side to C is BC and hypotenuse is CD

BC = Distance between the foot of the ladder and foot of the pole = m CD = length of the ladder = m

Exercise – 12.1 8. A boat has to cross a river. It crosses the river by making an angle of 600 with the bank of the river due to the stream of the river and travel a distance of 600 m to reach the another side of the river . What is the width of the river ? AB = Width of the river AC= Distance travelled by boad to reach another side of the river = 600 m BAC= The angle made by boat to crosses river with bank of the river= 600 600 Ò$. In ΔABC, Opposite side to A is BC, Adjacent side to A is AB and hypotenuse is AC AB= Width of the river = 300 m

BEA = The angle of elevation of the tree from observer eye = 450 Exercise – 12.1 9. An observer of height 1.8 m is 13.2 m away from a palm tree . The angle of elevation of the tree from his eyes is 450 . What is the height of the palm tree ? AC = AB + CB = height of the palm tree DE =CB = observer height = 1.8 m CD = Distance between Palm tree and observer = 13.2 m =BE BEA = The angle of elevation of the tree from observer eye = 450 In ΔABE, Opposite side to E is AB, Adjacent side to E is BE and hypotenuse is AE . 13.2 m 1.8 m 1.8 m AC = AB + CB =13.2+1.8=15 m 13.2 m height of the palm tree = 15 m

10. In the adjacent figure AC= 6 cm. AB=5 cm and BAC = 300 10. In the adjacent figure AC= 6 cm. AB=5 cm and BAC = 300 . Find the area of the triangle . Exercise – 12.1 Solution : Given that in the adjacent figure AC= 6 cm. AB=5 cm and BAC = 300 5 m . In ΔABC, Opposite side to A is BC, Adjacent side to A is AB and hypotenuse is AC. 6 m ΔABC Area= = 7.5 Sq. m