Question

Find the values of the six trigonometric functions of an angle in standard position if the point with coordinates (6,-5) lies on its terminal side.

Answer 1

Answer:

Given : The point with coordinates (6,-5) lies on its terminal side.

To find :  The values of the six trigonometric functions of an angle in standard position

Solution :

The coordinates (6,-5) lies in the 4 quadrant.

Refer the attached figure.

where x=6 is the base

y= -5 is the perpendicular

Applying Pythagoras theorem,

$H^2=P^2+B^2$

$H^2=(-5)^2+6^2$

$H^2=25+36$

$H=\sqrt{61}$

Now, finding trigonometric function

Note - In fourth quadrant only cos and sec is positive rest are negative.

1 )  $\sin x= - \frac{P}{H}$

$\sin x= - \frac{-5}{\sqrt{61}}$

$\sin x= \frac{5}{\sqrt{61}}$

2) $\csc x=\frac{1}{\sin x}$

$\csc x=\frac{1}{\frac{5}{\sqrt{61}}}$

$\csc x=\frac{\sqrt{61}}{5}$

3) $\cos x= \frac{B}{H}$

$\cos x=\frac{6}{\sqrt{61}}$

4) $\sec x=\frac{1}{\cos x}$

$\sec x=\frac{1}{\frac{6}{\sqrt{61}}}$

$\sec x=\frac{\sqrt{61}}{6}$

5) $\tan x= - \frac{P}{B}$

$\tan x= - \frac{-5}{6}$

$\tan x= \frac{5}{6}$

6) $\cot x=\frac{1}{\tan x}$

$\cot x=\frac{1}{\frac{5}{6}}$

$\cot x=\frac{6}{5}$

Answer 2

This is an angle in the 4th quadrant
tan 0f this angle = opp/adj = -5/6
cotangent = 1/tan = -6/5

length of hypotenuse  = sqrt (6^2 + (-5)^2 ) =  sqrt 61

so sine = -5/sqrt61
and cosec = 1/sin =  -aqrt61/5

and cosine = 6/sqrt61
and secant = 1 / cos =  sqrt61 /6