60 likes | 181 Views
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°.
E N D
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°
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°−
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°−
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°−
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°