Question

If sinA=4/3 find (2sinA-3cosA) /(2sinA+3cosA)

Answer 1

Answer:

Step-by-step explanation:

Given,

tanA=4/3

i.e, sinA/cosA = 4/3

so, p/b=4/3(p is perpendicular and b is base)

p=4 and b=3

so by pythagorous theorem,

h = root. p^2 + b^2

so, h =root. (4)^2 + (3)^2

so, h = root 16 +9

h= root. 25 =5

so,

2 sinA - 3 cosA/2sinA + 3cosA

= 2 p/h - 3 b/h/2 p/h + 3 b/h

= 2x 4/5 - 3x 3/5÷ 2x 4/5 + 3x 3/5

= 8/5 - 9/5÷ 8/5 + 9/5

= -1/5 ÷ 17/5

= -1/17

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