Question

Let (6,-5) be a point on the terminal side of θ. Find the exact values of sin θ, sec θ, and tan θ.

Answer 1

Answer:

sinθ = -5/√61

secθ = √61/6

tanθ = -5/6

Step-by-step explanation:

From the given coordinate (6, -5), x = 6 and y = -5. This shows that the point lies in the 4th quadrant. In the fourth quadrant, only cos θ is positive, both sin θ and tan θ are negative.

Let us get the value of the radius 'r' first before calculating the trigonometry identities.

Using the Pythagoras theorem;

$r^2 = x^2 + y^2\\r^2 = 6^2 + (-5)^2\\r^2 = 36+25\\r^2 = 61\\r = \sqrt{61}$

Using SOH, CAH, TOA to get the trigonometry identities;

Given x = 6, y =5 and r = √61

$sin \theta = opp/hyp = y/r\\sin \theta = 5/\sqrt{61}$

Since sin θ, is negative in the fourth quadrant, $sin \theta = -5/\sqrt{61}$

$cos \theta = adj/hyp = x/r\\cos \theta = 6/\sqrt{61}$

$sec \theta = \frac{1}{cos \theta} \\sec \theta = \frac{1}{6/\sqrt{61} } \\sec \theta = \frac{\sqrt{61} }{6}$

For tanθ:

$tan \theta = \frac{opp}{adj} = \frac{y}{x} \\tan \theta = \frac{5}{6}$

Since tan θ, is negative in the fourth quadrant, $tan \theta = -5/6$

Answer 2

 $SIN\Theta=\frac{-5}{\sqrt{61} }$,

$SEC\Theta=\frac{\sqrt{61} }{6}$  

$TAN\Theta=\frac{-5}{6}$.

Given coordinate of point be (6, -5).

So the point lies in the 4th quadrant. In the fourth quadrant, only cos θ is positive, both sin θ and tan θ are negative.

Since (6, -5) is on the terminal side that puts us in the fourth quadrant. If we plot that point and draw a diagonal to it from the origin, we will be able to draw a right triangle with that diagonal as the hypotenuse.That means the adjacent side is 6 and opposite side it -5.

By Pythagorean theorem, we find that the hypotenuse

$H=\sqrt{6^{2} +(-5)^{2} }$

$H=\sqrt{61}$

We know from trigonometric ratios,

$SIN\Theta=\frac{Perpendicular}{hypotenuse}$

$SIN\Theta=\frac{5}{\sqrt{61} }$

Since $SIN\Theta$ is negative in 4th quadrant,

so,

$SIN\Theta=\frac{-5}{\sqrt{61} }$

Similarly,

$SEC\Theta=\frac{hypotenuse}{base}$

$SEC\Theta=\frac{\sqrt{61} }{6}$

And,

$TAN\Theta=\frac{perpendicular}{base}$

$TAN\Theta=\frac{-5}{6}$

Hence,  $SIN\Theta=\frac{-5}{\sqrt{61} }$, $SEC\Theta=\frac{\sqrt{61} }{6}$  and $TAN\Theta=\frac{-5}{6}$.

For more details follow the link:

https://brainly.com/question/1201366