Answer:
[tex]\sin\alpha=\dfrac{3}{\sqrt{13}}[/tex]
Step-by-step explanation:
We have been given the relation:-
[tex]\dfrac{\sin\alpha+cos\alpha}{\sin\alpha-\cos\alpha}=5[/tex]
Cross multiply the equation:-
[tex](\sin\alpha+\cos\alpha)=5(\sin\alpha-\cos\alpha) \\ \\\sin\alpha+\cos\alpha=5\sin\alpha-5\cos\alpha \\ \\6\cos\alpha=4\sin\alpha \\ \\\tan\alpha = \dfrac{3}{2}[/tex]
The tangent function for a triangle is the ratio of Perpendicular over Base.
For a triangle, with its Base as 2 units and Perpendicular as 3 units.
This triangle will have its Hypotenuse as:-
[tex]H = \sqrt{{Base}^{2}+{Perpendicular}^{2}} \\ \\H = \sqrt{3^2+2^2} \\ \\H = \sqrt{13} \ units[/tex]
The Sine function for a triangle is the ratio of Perpendicular over Hypotenuse.
Therefore,
[tex]\sin\alpha=\dfrac{Perpendicular}{Hypotenuse} \\ \\ \sin\alpha=\dfrac{3}{\sqrt{13}}[/tex]
