1. Find csc A, cos A and tan A of an angle A whose terminal side contains (– 12, – 5).
2. Find sin X, sec X and cot X of an angle X whose terminal side contains (– 3, 5).
You know that
sinØ = y/r
cosØ = x/r
tanØ = y/x
cscØ = r/y
etc
so in any point the first coordinate is the x, the 2nd is the y
1. for (-12, -5) you would be in quadrant III
x = -12, y = -5
use Pythagoras, x^2 + y^2 = r^2 ( where we keep r always positive)
we get r = 13
now simply form the ratios
sinØ = y/r = -5/13
etc
Do the same thing for #2
To solve these problems, we can use the Pythagorean Identity and the definition of trigonometric functions.
1. Find csc A, cos A, and tan A of an angle A whose terminal side contains (-12, -5):
Step 1: Find the hypotenuse (r) using the distance formula:
r = √((-12)^2 + (-5)^2)
= √(144 + 25)
= √169
= 13
Step 2: Use the given coordinates to determine the trigonometric ratios:
csc A = r / y = 13 / -5 = -13/5
cos A = x / r = -12 / 13
tan A = y / x = -5 / -12 = 5/12
Therefore, csc A = -13/5, cos A = -12/13, and tan A = 5/12.
2. Find sin X, sec X, and cot X of an angle X whose terminal side contains (-3, 5):
Step 1: Find the hypotenuse (r) using the distance formula:
r = √((-3)^2 + 5^2)
= √(9 + 25)
= √34
Step 2: Use the given coordinates to determine the trigonometric ratios:
sin X = y / r = 5 / √34
sec X = r / x = √34 / -3 = -√34/3
cot X = x / y = -3 / 5
Therefore, sin X = 5/√34, sec X = -√34/3, and cot X = -3/5.