2 years ago

3x^2-5=0 completing square

Mathematics

1 answer:

Valentin [98]2 years ago

5 0

It would be good if you add the answers choice, but I guess it’s (b/2)^2

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Assume that the terminal side of thetaθ passes through the point (negative 12 comma 5 )(−12,5) and find the values of trigonomet

zmey [24]

Answer:

$\sin \theta = \dfrac{5}{13}$ and $\sec \theta = -\dfrac{13}{12}$

Step-by-step explanation:

Assume that the terminal side of thetaθ passes through the point (−12,5).

In ordered pair (-12,5), x-intercept is negative and y-intercept is positive. It means the point lies in 2nd quadrant.

Using Pythagoras theorem:

$hypotenuse^2=perpendicular^2+base^2$

$hypotenuse^2=(5)^2+(12)^2$

$hypotenuse^2=25+144$

$hypotenuse^2=169$

Taking square root on both sides.

$hypotenuse=13$

In a right angled triangle

$\sin \theta = \dfrac{opposite}{hypotenuse}$

$\sin \theta = \dfrac{5}{13}$

$\sec \theta = \dfrac{hypotenuse}{adjacent}$

$\sec \theta = \dfrac{13}{12}$

In second quadrant only sine and cosecant are positive.

$\sin \theta = \dfrac{5}{13}$ and $\sec \theta = -\dfrac{13}{12}$

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3 years ago

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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