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Alex_Xolod [135]
3 years ago
11

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

ric ratios sec thetaθ and sin thetaθ.

Mathematics
1 answer:
zmey [24]3 years ago
6 0

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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