Question

The terminal side of angle B in standard position goes through the point (-6,-9). Find the values of the six trigonometric functions of B. Give an exact answer in reduced radical form. No decimal answer will be accepted!! sin(B) = cos(B) = tan(8) csc(B) sec() = cot(B) =

Answer 1

Sal? Let's draw a triangle (suight angled) having an angle B. o point be the origin (0,0) Al-6,-9) Point A be (-6,-9) ie 'A' point 9 units is in Il quadrant. JB B oc 6 units Using Pythagoras theorom in AAOB (AB)2 + (30)? (AO)? - (932 + 6002 (AO)? 81 + 36 AO = 5117 AO = 3/13 -3 sin(B) = AB AO -9 3513 un negative as only tanß & cot B is positive in TII quadrant - 3813 cos (B) OB AO -6 3 53 sinß - sin 13 Cos ß = -2513 13 tan ß = AB OB oto tan ß = 3. 2

cot ß OB AB do cot ß = 2 3 - 353 sec ß - cas ß 6 sec ß- 113 2 CSC ß- - 13 1 sinf 303 CSc ß = -√3 3