Answer:
Step-by-step explanation:
tan Θ + tan 2Θ + √3 tan Θ tan 2Θ = √3
tan Θ + tan 2Θ = √3 - √3 tan Θ tan 2Θ
tan Θ + tan 2Θ = √3 ( 1 - tan Θ tan 2Θ)
(tan Θ + tan 2Θ) / (1 - tanΘ tan 2Θ) = √3
tan(Θ + 2Θ) = √3
tan 3Θ = tan (
) we know tan Θ = tan α; Θ = nΠ + α, n belongs to z
3Θ = nΠ + Π/3
Θ = nπ/3 + Π/9 for all n in Z
I don't know what the "six-step method" is supposed to be, so I'll just demonstrate the typical method for this problem.
Let <em>x</em> be the amount (in gal) of the 50% antifreeze solution that is required. The new solution will then have a total volume of (<em>x</em> + 60) gal.
Each gal of the 50% solution used contributes 0.5 gal of antifreeze. Similarly, each gal of the 30% solution contributes 0.3 gal of antifreeze. So the new solution will contain (0.5 <em>x</em> + 0.3 * 60) gal = (0.5 <em>x</em> + 18) gal of antifreeze.
We want the concentration of antifreeze to be 40% in the new solution, so we need to have
(0.5 <em>x</em> + 18) / (<em>x</em> + 60) = 0.4
Solve for <em>x</em> :
0.5 <em>x</em> + 18 = 0.4 (<em>x</em> + 60)
0.5 <em>x</em> + 18 = 0.4 <em>x</em> + 24
0.5 <em>x</em> - 0.4 <em>x</em> = 24 - 18
0.1 <em>x</em> = 6
<em>x</em> = 6/0.1 = 60 gal
By doing a two step equation
