If A+B, A are acute angles such that sin (A + B) =24/25 and tan A =3/4,
then find the value of cos B
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Step-by-step explanation:
sin(A+B)=sinAcosB+cosAsinB....1
tanA=3/4=perpendicular/base
let perpendicular be 3x and base be 4x
hypo^2 = 9x^2+16x^2
hypo^2 = 25x^2
hypo = 5x
SinA = perpendicular/hypo = 3/5
cosA = base/hypo = 4/5
from eq1
24/25= 3cosB/5+4sinB/5
cos^2B+sin^2B=1
solve the two eq
hope it helps u
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