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