Step-by-step explanation:
3. We know the x value which is -5 and the y value which is 1. We need to find the hypotenuse which we refer to the r. We use the pythagorean theorem for that
[tex] {x}^{2} + {y}^{2} = {r}^{2} [/tex]
[tex] { - 5}^{2} + {1}^{2} = r {}^{2} [/tex]
[tex]25 + 1 = r {}^{2} [/tex]
[tex]26 = r {}^{2} [/tex]
[tex]r = \sqrt{26} [/tex]
Now let find the sin, csc, and tan.
To find sin let use the formula
[tex] \frac{y}{r} [/tex]
[tex] \frac{1}{ \sqrt{26} } [/tex]
Rationalize the denominator
[tex] \frac{1}{ \sqrt{26} } \times \frac{ \sqrt{26} }{ \sqrt{26} } = \frac{ \sqrt{26} }{26} [/tex]
[tex] \sin = \frac{ \sqrt{26} }{26} [/tex]
To find cosecant, it is the reciprocal of sin so it would be
[tex] \csc = \frac{26}{ \sqrt{26} } \times \frac{ \sqrt{26} }{ \sqrt{26} } = \sqrt{26} [/tex]
To find tan, do
[tex] \frac{y}{x} [/tex]
Which is simply
[tex] \tan= \frac{1}{ - 5} [/tex]
4. Kennedi is correct since in the 2nd quadrant
The x value is negative, and the y value is positive.
Since sin is referred in trig as the y value, sin must be positive.
