Answer with explanation:
⇒Position of point , which is on the terminal ray of Angle Theta = (5, -2)
⇒Let this point be Represented as P (5, -2) and Origin is Represented by Point O(0,0).
⇒Join OP and draw Perpendicular from P on the X axis,which cut the X axis at A(5,0).
This is Required Right Δ O AP.
⇒Find the Length of OP , using Distance formula.
[tex]D=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}\\\\OP=\sqrt{(5-0)^2+(-2-0)^2}\\\\OP=\sqrt{29}[/tex]
O A=5 units
PA=2 units
⇒As, the point lies in fourth Quadrant,Only Secant and Cosine function will be positive, all other trigonometric function will have negative value.
By Applying , trigonometric ratio formula,as we have length of three sides of triangle we can evaluate six trigonometric ratios.
The standard position of the terminal ray of an angle is [tex](x,y) = (5,-2)[/tex]
To determine the six trigonometric functions, we start by calculating the three basic functions using the following formulas
[tex]\sin(\theta) = \frac{y}{\sqrt{x^2 + y^2}}[/tex]
[tex]\cos(\theta) = \frac{x}{\sqrt{x^2 + y^2}}[/tex]
[tex]\tan(\theta) = \frac{y}{x}[/tex]
The other functions are the inverse of the above functions.
So, we have:
[tex]\csc (\theta) = \frac{\sqrt{x^2 + y^2}}{y}[/tex]
[tex]\sec (\theta) = \frac{\sqrt{x^2 + y^2}}{x}[/tex]
[tex]\cot(\theta) = \frac{x}{y}[/tex]
Read more about terminal ray at:
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