3. Angle of Elevation Ram Ravi )  Eye level Ram looks up above Ravi ‘  ’ Is the angle of elevation

4. Angle of depression  ( Eye level Ravi looks down at Ram Ravi Ram  Is the angle of ‘depression’

5. Angle of elevation Angle of depression Eye level ))) ))

6. If the object is above the horizontal level of the eyes we raise our head upwards to view the object. Our eyes move through an angle to view the object. The angle between the horizontal line and the line of sight is called the angle of elevation θ

7. If the object is below the horizontal level of the eyes we move down our eyes through an angle to view the object. The angle between the horizontal line and the line of sight is called the angle of depression. Ф

8. Steps in Solving problems To solve problems the first step is to draw the correct labeled diagram based on the question. Mark the sides and angles given in the question. Take the unknown quantities to be found out as ‘x’, ‘y’, ‘p’ etc, and mark in the diagram. Identify the correct trigonometric ratio which can be formed with the given quantities and the unknown quantity to be found out.

9. How will the man find the height of tip of the flag post from the ground? 1500m 45°(

10. Solution Let the height of the hill be ‘h’ m Tan 45° = h/1500 Or, 1 = h/1500  h = 1500 m 1500m 45°(

11. ø b h2h2 tan  =h 1 /a tan ø=h 2 /b a,b, ø &  can be measured to find h 1 & h 2  a h1h1

12. Hill, far away from the measuring point Take two points A,B (say) the distance between them is measurable(500m) The angle of elevation at A is (45°) the angle of elevation at B is (60°) Hill height= h BA 500m ) 60 ) 45

13. C D height= h B A 500m ) 60 o ) 45 o Tan45°=h/AD 1=h/AD  h=AD (or) h=AB + BD = 500+BD---(1) Also tan 60°=h/BD  3=h/BD BD=h/  3 -------(2) From (1) & (2), find ‘h’

14. h = 500+BD--- (1) BD = h/  3 -------(2) Substitute (2) in (1) h = 500 + h/  3 (or)  3h – h = 500  3 (or) h (  3 – 1 ) = 500  3 (or) h = 500  3 (  3 – 1 ) = 500  3 (  3 + 1 ) (  3 – 1 ) (  3+1 ) = (1500-500 x1.732)/2 Simplify & find the answer.

15. AB is light house C and D are positions of boats A man in a boat observes that angle of elevation changes from 60 to 45 in 2 minutes B A CD 45° 60° 150m150m

16. To Find Speed AD \ AB = cot 45 ° = 1 AD \ 150 = 1 AD = 150 m Let AC = X meters CD = (150 -x) AC \ AB = cot 60 ° = 1 \  3  x \ 150 = 1 \  3 A 45° 60° B 45° C D A 60° B

17. X = ( 150 \ \/3 )m = ( 50  3 ) m CD = (150 - 50  3) m = 63.4 m Boat covers 63.4 m in 2 minutes. Speed of boat = ( 63.4 ÷120 ) m\s = 0.53 m\s

18. To Find Height An observer is 1.6 m tall and is 45 meters away. The angle of elevation from his eye to top of tower is 30  Determine height of tower. 30° 45m 1.6m

19. Solution H eight of the tower = h+1.6m h=45 tan30° h=45 X 1/  3 h=45 X  3/3 =15X1.73 =25.95  Height=27.55 m 1.6 30° 45m h

20. From the top of a tower 50m high the angles of depression of the top and bottom of a pole are observed to be 45º and 60 º respectively. Find the height of the pole, and the tower stand in the same line. towertower polepole 50m 60° 45 °

21. Solution A tower polepole 50m 60° 45 °AB is the tower B C D CD is the pole 60° 45 °

22. EB=CD & EC =BD Let Height of the pole CD =h m Take  AEC &  ABD to solve for “h” 50m B C D 60° 45 ° A E h

23. There is a small island in the middle of a 100m wide river and a tall tree stands on the island. P and Q are points directly opposite each other on the two banks, and in line with the tree. If the angles of elevation of the top of the tree from P and Q are respectively30 º and 45º,find the height of the tree. 100 m PQ 30° 45°

24. Tree height =h m (say) h/PR =tan30° =1/  3  3 h =PR  3 h =100-RQ………..(1) h/RQ =tan45° =1 h =RQ………..(2) There is a small island in the middle of a 100m wide river and a tall tree stands on the island. P and Q are points directly opposite each other on the two banks, and in line with the tree. If the angles of elevation of the top of the tree from P and Q are respectively30 º and 45º,find the height of the tree. treetree P Q 30 45 R

25. From (1) & (2)  3 h =100-h h(  3 -1) =100 H =100/(  3 -1) = 100 (  3 +1) /(  3 -1) /(  3 +1) 100x2.732/2 27.32/2 13.66 m Height of the tree =13.7 m

26. A θ θθ θ ф h’ h W P From a window ( h meters high above the ground) of a house in a street,the angle of elevation and depression of the top and the foot of another house on opposite side of the street are θ and ф respectively. show that the height of the opposite house ish(1+tan θ cot ф ). street househouse

27. Let W be the window and AB be the house on the opposite side. then WP is the width of the street. In ΔBPW, tan ф =PB/WP h/WP = tan ф or WP = h cot ф In Δ AWP, tan θ = AP/WP or h’/WP = tan θ or h’ = WP tan θ h’ = h cot ф tan θ Height of the house = h + h’ =h+ h tan θ cot ф =h(1+ tan θ cot ф )