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TRIGONOMETRY

TRIGONOMETRY. KALPALATHA GHS SANTHIPURAM. TRIGONOMETRY. SOGADABALLA. NAME OF THE SIEDES OF RIGHTANGLE TRIANGLE. C. HYPOTENUSE. OPPOSITESIDE. B. A. ADJESENTSIDE. C. Sin θ =. HYPOTENUSE. HYPOTENUSE. OPPOSITESIDE. OPPOSITESIDE. θ. A. A. B. ADJESENTSIDE. ADJESENTSIDE. C.

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TRIGONOMETRY

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  1. TRIGONOMETRY KALPALATHA GHS SANTHIPURAM

  2. TRIGONOMETRY SOGADABALLA

  3. NAME OF THE SIEDES OF RIGHTANGLE TRIANGLE C HYPOTENUSE OPPOSITESIDE B A ADJESENTSIDE

  4. C Sinθ= HYPOTENUSE HYPOTENUSE OPPOSITESIDE OPPOSITESIDE θ A A B ADJESENTSIDE ADJESENTSIDE

  5. C Cosθ= HYPOTENUSE HYPOTENUSE OPPOSITESIDE OPPOSITESIDE θ A A ADJESENTSIDE ADJESENTSIDE B

  6. C Cosecθ= HYPOTENUSE HYPOTENUSE OPPOSITESIDE OPPOSITESIDE θ A A B ADJESENTSIDE ADJESENTSIDE

  7. C Secθ= HYPOTENUSE HYPOTENUSE OPPOSITESIDE OPPOSITESIDE θ A A ADJESENTSIDE ADJESENTSIDE B

  8. C Tanθ= HYPOTENUSE OPPOSITESIDE OPPOSITESIDE θ A ADJESENTSIDE ADJESENTSIDE B

  9. C Cotθ= HYPOTENUSE OPPOSITESIDE OPPOSITESIDE θ A ADJESENTSIDE ADJESENTSIDE B

  10. 0º 30º 45º 60º 90º Sinθ 1/2 1/√2 1/√2 √3/2 √3/2 1 1 1/2 0 0 Cosθ Tanθ 0 0 1/√3 1/√3 √3 √3 ∞ ∞ 1 1 2/√3 2/√3 Cscθ √2 √2 ∞ ∞ 2 2 1 1 Secθ Cotθ

  11. OBJECT ANGLE OF ELEVATION LINE SIGHT OBSERVAR HORIZONTAL

  12. ANGLE OF DEPRESSION C HORIZANTAL Angle of depression LIGHT OF SIGHT B A HORIZANTAL OBJECT

  13. 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. 21°.48' 12°.30'

  14. C 12°.30' 21°.48' 12°.30' D E 21°.48' B A 90°

  15. 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 C 12°.30' D E 21°.48' B A

  16. 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.

  17. A 52°.10' 41°.20' 2500mts 52°.10' D 41°.20' B C

  18. 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 A 2500mts C B E

  19. ACKNOWLEDGEMENT 10thCLASS TEXT BOOK 9th CLASS TEXT BOOK 8th CLASS TEXT BOOK www.animatononline.com

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