Answer:
[tex]\textsf{a)}\quad y = \dfrac{4}{5}[/tex]
[tex]\textsf{b)}\quad \cos \theta = \dfrac{3}{5}[/tex]
[tex]\textsf{c)}\quad \sin \theta = \dfrac{4}{5}[/tex]
[tex]\textsf{d)}\quad \tan \theta = \dfrac{4}{3}[/tex]
Step-by-step explanation:
We are given that P is the point at which the terminal side of an angle θ in standard position intersects the unit circle, and that P has the coordinates (3/5, y) in the first quadrant.
[tex]\dotfill[/tex]
Part a
Since P lies on the unit circle, we can use the equation of the unit circle x² + y² = 1 to find the y-coordinate of P.
Substitute x = 3/5 into the equation and solve for y:
[tex]\left(\dfrac{3}{5}\right)^2 + y^2 = 1\\\\\\\dfrac{9}{25} + y^2 = 1\\\\\\y^2=1-\dfrac{9}{25}\\\\\\y^2=\dfrac{25}{25}-\dfrac{9}{25}\\\\\\y^2=\dfrac{25-9}{25}\\\\\\y^2=\dfrac{16}{25}\\\\\\y=\pm \sqrt{\dfrac{16}{25}}\\\\\\y=\pm \dfrac{4}{5}[/tex]
Since the point P is in the first quadrant, y must be positive, so we take the positive square root:
[tex]y = \dfrac{4}{5}[/tex]
[tex]\dotfill[/tex]
Part b
The cosine of the angle θ represents the x-coordinate of a point on the unit circle. As the x-coordinate of P is x = 3/5 then:
[tex]\cos \theta = \dfrac{3}{5}[/tex]
[tex]\dotfill[/tex]
Part c
The sine of the angle θ represents the y-coordinate of a point on the unit circle. As the y-coordinate of P is y = 4/5 then:
[tex]\sin \theta = \dfrac{4}{5}[/tex]
[tex]\dotfill[/tex]
Part d
The tangent of an angle is the ratio of the sine and cosine of that angle. Therefore:
[tex]\tan \theta = \dfrac{\sin \theta}{\cos \theta}\\\\\\\tan \theta = \dfrac{\frac45}{\frac35}\\\\\\\tan \theta = \dfrac{4}{3}[/tex]
