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

Guest Oct 15, 2014

+23254

+5

Best Answer

sin(A) = 12/13 in quadrant 2 tan(B) = -4/3 in quadrant 4

Since A is in Q2, the missing side is -5 and cos(A) = -5/13 and tan(A) = -12/5

tan(A+B) = (tan(A) + tan(B) / (1 - tan(A)tan(B))

= (-12/5 + -4/3) / (1 - (-12/5)(-4/3)) = 56/33

geno3141 Oct 15, 2014

geno3141 · Answer 1 · 2014-10-15T06:55:46

sin(A) = 12/13 in quadrant 2 tan(B) = -4/3 in quadrant 4

Since A is in Q2, the missing side is -5 and cos(A) = -5/13 and tan(A) = -12/5

tan(A+B) = (tan(A) + tan(B) / (1 - tan(A)tan(B))

= (-12/5 + -4/3) / (1 - (-12/5)(-4/3)) = 56/33