Question

Angle θ is in standard position. If (8, -15) is on the terminal ray of angle θ, find the values of the trigonometric functions.

Answer 1

$\bf \begin{array}{rlclll} (8&,&-15)\\ \uparrow &&\uparrow \\ a&&b \end{array} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{8^2+(-15)^2}\implies c=\sqrt{289}\implies \boxed{c=17}\\\\ -------------------------------$

$\bf sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad cos(\theta)=\cfrac{adjacent}{hypotenuse} \quad % tangent tan(\theta)=\cfrac{opposite}{adjacent} \\\\\\ % cotangent cot(\theta)=\cfrac{adjacent}{opposite} \qquad % cosecant csc(\theta)=\cfrac{hypotenuse}{opposite} \quad % secant sec(\theta)=\cfrac{hypotenuse}{adjacent}$

you've got all three values, plug them in.

Answer 2

Answer with explanation:

It is given that, Angle  θ is in standard position.

A line from origin O to point , P(8,-15) is joined and then perpendicular to x and y axis, is drawn cutting X axis at Point M and Y axis at point N.

OM= 8 units

ON=P M=15 units

By Pythagorean Theorem

$OM^2 + PM^2=OP^2\\\\ 8^2 +15^2=OP^2\\\\ OP^2=64 +225\\\\OP^2=289\\\\OP^2=17^2\\\\OP=17\\\\ Sin (\theta)=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{-15}{17}\\\\Cos(\theta)=\frac{\text{Base}}{Hypotenuse}=\frac{8}{17}\\\\Tan(\theta)=\frac{\text{Perpendicular}}{Base}=\frac{-15}{8}\\\\ Cosec(\theta)=\frac{1}{Sin(\theta)}=\frac{-17}{15}\\\\ Sec(\theta)=\frac{1}{Cos(\theta)}=\frac{17}{8}\\\\ Cot (\theta)=\frac{1}{Tan(\theta)}=\frac{-8}{15}$

Point(8,-15), lies in Quadrant four. In Quadrant four Cosine and Secant Function are positive and all other trigonometric functions, Sine,Cosecant, Tangent,and Cotangent are Negative.