1 / 6

TRIGONOMETRY

TRIGONOMETRY. . Sign for sin  , cos  and tan . Quadrant II 90 ° <  < 180°. SIN  (+). ALL (+). Quadrant I 0 ° <  < 90°.  = 180 °− . Let  = acute angle.  = . . . . .  = 180 °+ .  = 360 °− . TAN  (+). COS  (+). Quadrant IV 270 ° <  < 360°.

mira-meyers
Télécharger la présentation

TRIGONOMETRY

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. TRIGONOMETRY

  2. Sign for sin , cos  and tan  Quadrant II 90° <  < 180° SIN  (+) ALL (+) Quadrant I 0° <  < 90°  = 180°−  Let  = acute angle  =       = 180°+   = 360°−  TAN  (+) COS  (+) Quadrant IV 270° <  < 360° Quadrant III 180° <  < 270°

  3. Finding angle  when given sin  Quadrant I 0° <  < 90°  =  Quad I & Quad II sign (+) Given that 0°   360°, find  when sin  = 0.7660 sin  = −0.5736 • = sin-1 0.7660 • = 50° (acute angle)   = 50°, 130° Quadrant II 90° <  < 180° SIN  (+)  = 180°−  Quad III & Quad IV sign (−) Quadrant III 180° <  < 270° • = sin-1 0.5736 • = 35°   = 180° + 35°, 360°−35° = 215°, 325° TAN  (+)  = 180°+  Quadrant IV 270° <  < 360° COS  (+)  = 360°− 

  4. Finding angle  when given cos  Quadrant I 0° <  < 90°  =  Quad I & Quad IV sign(+) Given that 0°   360°, find  when • cos  = 0.7660 • cos  = −0.5736 • = cos-1 0.7660 • = 40°   = 40°, 360 − 40° = 40°, 320° Quadrant 2 90° <  < 180° SIN  (+)  = 180°−  Quadrant 3 180° <  < 270° Quad II & Quad III sign (−) TAN  (+) • = cos-1 0.5736 • = 55°   = 180° −55°, 180°+35° = 125°, 235°  = 180°+  Quadrant 4 270° <  < 360° COS  (+)  = 360°− 

  5. Find angle  when given tan  Quadrant 1 0° <  < 90°  =  Quadrant I and Quadrant 3 sign (+) Given that 0°   360°, find  when • tan  = 1.7660 • tan  = −2.5 • = tan-1 1.7660 • = 60°29’ Hence  = 60°29’, 180° + 60°29’ = 60°29’, 240° 29’ Quadrant 2 90° <  < 180° SIN  (+)  = 180°−  Quadrant 2 and Quadrant 4 Quadrant 3 180° <  < 270° sign (−) TAN  (+) • = tan-1 2.5 • = 68°12’ Hence  = 180° − 68°12’, 360°−68°12’ = 111°48’, 291°48’  = 180°+  Quadrant 4 270° <  < 360° KOS  (+)  = 360°− 

  6. Practice makes perfect!!! 1. Given sin x° =0.7547 and 90° x  180°, find x. 2. Given cos x = cos 34° and 270° x  360°, find x. 3. Given cos x = − 0.6926 and 90° x 180°, find x. 4. Given tan x = 0.8 and 180° x  360°, find x. 5. Given tan x = −0.8098 and 270° x  360°, find x. Answer: (1) 131° (2)326° (3)133°50’ (4)218°40’ (5)321°

More Related