Trig
posted by Anonymous .
Let sin A = 12/13 with 90º≤A≤180º and tan B = 4/3 with 270º≤B≤360º. Find tan (A + B).

make quick sketches of rightangled triangles for each of the two cases, and use Pythagoras to find the missing sides.
first triangle:
since sinA = 12/13 , y = 12 and r = 13
x^2 + y^2 = r^2
x^2 + 144 = 169
x^2 = 25
x = ±5, but we are in quad II, so x = 5
then tanA =  12/5
2nd triangle:
since tanB = 4/3
tan(AB) = (tanA  tanB)/(1 + tanAtanB)
= (12/5  (4/3) )/( 1 + (12/5)(4/3)
= (16/15) / (21/5)
= 16/63 
Are you sure it's negative?
Respond to this Question
Similar Questions

algebra 1 help please
4) a student score is 83 and 91 on her first two quizzes. write and solve a compound inequality to find possible values for a thord quiz score that would give anverage between 85 and 90. a. 85≤83+91+n/3 ≤90; 81≤n≤96 … 
trig
Let sin A = 12/13 with 90º≤A≤180º and tan B = 4/3 with 270º≤B≤360º. Find tan (A  B). 
Trig
Suppose sinA = 12/13 with 90º≤A≤180º. Suppose also that sinB = 7/25 with 90º≤B≤0º. Find cos(A  B). 
Trig
Suppose sin A = 12/13 with 90º≤A≤180º. Suppose also that sin B = 7/25 with 90º≤B≤0º. Find tan (A  B). 
Math
Suppose sinA = 12/13 with 90º≤A≤180º. Suppose also that sinB = 7/25 with 90º≤B≤0º. Find cos(A + B). 
Math
Suppose sinA = 12/13 with 90º≤A≤180º. Suppose also that sinB = 7/25 with 90º≤B≤0º. Find cos(A  B). 
Math
Let sin A = 12/13 with 90º≤A≤180º and tan B = 4/3 with 270º≤B≤360º. Find tan (A + B). 
Trigonometry
Suppose sin A = 12/13 with 90º≤A≤180º. Suppose also that sin B = 7/25 with 90º≤B≤0º. Find tan (A – B). 
Math
Suppose cosA = 12/13 with 0º≤A≤90º. Suppose also that sinB = 8/17 with 90º≤B≤180º. Find cos(A  B). 
PRE  CALCULUS
Eliminate the parameter t. Find a rectangular equation for the plane curve defined by the parametric equations. x = 6 cos t, y = 6 sin t; 0 ≤ t ≤ 2π A. x2  y2 = 6; 6 ≤ x ≤ 6 B. x2  y2 = 36; 6 ≤ …