2. TRIGONOMETRY KALPALATHA GHS SANTHIPURAM

3. SOGADABALLA

4. O P P O S I T E S I D E ADJESENTSIDE H Y P O T E N U S E A C B NAME OF THE SIEDES OF RIGHTANGLE TRIANGLE

5. A OPPOSITESIDE ADJESENTSIDE H Y P O T E N U S E Sin θ= HYPOTENUSE OPPOSITESIDE ADJESENTSIDE θ

6. A O P P O S I T E S I D E ADJESENTSIDE H Y P O T E N U S E Cos θ= HYPOTENUSE O P P O S I T E S I D E ADJESENTSIDE θ

7. A OPPOSITESIDE ADJESENTSIDE H Y P O T E N U S E Cosec θ= HYPOTENUSE OPPOSITESIDE ADJESENTSIDE θ

8. A O P P O S I T E S I D E ADJESENTSIDE H Y P O T E N U S E Sec θ= HYPOTENUSE O P P O S I T E S I D E ADJESENTSIDE θ

9. O P P O S I T E S I D E H Y P O T E N U S E Tan θ= OPPOSITESIDE ADJESENTSIDE θ

10. O P P O S I T E S I D E H Y P O T E N U S E Cot θ= OPPOSITESIDE ADJESENTSIDE θ

11. Cotθ Secθ Cscθ Tanθ Cosθ Sinθ 90º60º45º30º0º 1 1/ √2 0 0 1/2√3/2 0 11/ √2 √3/2 1/2 1 1/ √3 √3 ∞ ∞∞ 22 √ 2 2/ √ 3 2/ √ 3 1 1 √ 2 ∞ 1 1/ √3 0

12. ANGLE OF ELEVATION OBJECT HORIZONTAL OBSERVAR LINE SIGHT

13. ANGLE OF DEPRESSION HORIZANTAL LIGHT OF SIGHT HORIZANTALOBJECT Angle of depression

14. There are two temples, one on each bank of the river, just opposite to each Other.one of the temples A is 40 mts high.As observed from the top of this temple A, the angle of depression of the top and foot of the other temple B are and respectively. Find the width of the river and the height of the temple B approximately. 12°.30' 21°.48'

15. 12°.30' 12°.30' 21°.48' 21°.48' 90°

16. E D C B A 21°.48' 12°.30' AB= river AC=BD=temples C=observer In right angle triangle ABC TanB=AC/AB Tan 21°.48' =40/AB 0.400 =40/AB AB =40/0.4 AB =100 mts In right angle triangle CED TanD=CE/ED Tan12°.30‘=CE/100 CE =0.2217×100=22.17 mts Height of the temple AC=AE+EC 40 =BD+22.17 ( BD=AE) BD =40-22.17=17.83 mts Height of the B temple =17.83 mts

17. An aeroplane at an altitude of 2500 mts. Observes the angles of depression of opposite points on the two banks of river to be 41°.20' and 52°.10'. Find the width of the river in metres.

18. 41°.20' 52°.10' 41°.20' 52°.10' 2500mts

19. A BC E AB=2500mts C and D are two bank river. CD is the width of the river Given ∟ACB= 41°.20‘ and ∟ ADB= 52°.10‘ From right angle triangle ACB tan 41°.20‘ =AB/CB 0.8795=2500/CB CB = 2500/0.8795 = 2842.5mts from right angle triangle ABD, ∟ADB = 52°.10‘ Tan 52°.10‘ = AB/BD 1.2876 = 2500/BD BD =2500/1.2876=1941.5mts CD=CB+BD =2842.5+1941.5=4784 mts The width of the river =4784 mts

20. 10th CLASS TEXT BOOK 9 th CLASS TEXT BOOK 8 th CLASS TEXT BOOK www.animatononline.com