Mathematics, 18.08.2020 17:01 caroline1484

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

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Answer from: ozzy1146

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

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

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

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

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Answer from: angrybirdsryver

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 from: Quest

answer:

3 bags of apples and 2 bags of oranges.

step-by-step explanation:

Answer from: Quest

x =-.20752

step-by-step explanation:

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know logb c = loga c/loga b

we want to convert to base 10 so a = 10

by default, we do not write the 10 c = 4 and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752

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

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