Please tell

Asked by  | 20th Jan, 2009, 09:08: PM

Expert Answer:

3sinA - 4cosA = 5

(3/5)sinA - (4/5)cosA = 1

let cosB = 3/5 then sinB = 4/5

cosBsinA - sinBcosA = sin 90

sin(A-B) = sin 90

A-B = 90

A = 90+B

now 3cosA+4sinA

=3cos(90+B) + 4sin(90+B)

=3(-sinB) + 4cosB

=3(-4/5) + 4(3/5)

=0

Answered by  | 22nd Jan, 2009, 10:21: AM

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