Question

If 4sinθ = `sqrt(13)`, find the value of `(4sinθ - 3cosθ)/(2sinθ + 6cosθ)`

Answer 1

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)`.