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Question
If 4sinθ = `sqrt(13)`, find the value of `(4sinθ - 3cosθ)/(2sinθ + 6cosθ)`
Solution
Consider ΔABC, where ∠B = 90°
⇒ 4sinθ = `sqrt(13)`
⇒ 4sinθ = `sqrt(13)/(4)`
⇒ sinθ = `"Perpendicular"/"Hypotenuse" = "BC"/"AC" = sqrt(13)/(4)`
By Pythagoras theorem,
AC2 = AB2 + BC2
⇒ AB2
= AC2 - BC2
= `4^2 - (sqrt(13))^2`
= 16 - 13
= 3
⇒ AB = `sqrt(3)`
Now,
cosθ = `"Base"/"Hypotenuse" = "AB"/"AC" = sqrt(3)/(4)`
`(4sinθ - 3cosθ)/(2sinθ + 6cosθ)`
= `(4 xx sqrt(3)/(4) - 3 xx sqrt(3)/4)/(2 xx sqrt(13)/4 + 6 cc sqrt(3)/4)`
= `((4sqrt(13) - 3sqrt(13))/(4))/((2sqrt(13) + 6sqrt(13))/(4)`
= `(4sqrt(13) - 3sqrt(13))/(2sqrt(13) + 6sqrt(13)`
= `sqrt(13)/(8sqrt(13)`
= `(1)/(8)`.
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