The coordinates of a point P on the terminal arm of each angle are shown. Write the exact trigonometric ratios sin θ, cos θ, and tan θ for each.
Problem 1 :
Solution :
By drawing a perpendicular line from P to the x-axis, we will get a right triangle. Then OP be the hypotenuse.
Use the Pythagorean Theorem,
OP = √(3² + 4²) OP = √(9 + 16) OP = √25 OP = 5 |
Opposite side = 4 Adjacent side = 3 Hypotenuse = 5 |
sin θ = Opposite side / Hypotenuse
sin θ = 4/5
cos θ = Adjacent side / Hypotenuse
cos θ = 3/5
tan θ = Opposite side / Adjacent side
tan θ = 4/3
Problem 2 :
Solution :
By drawing a perpendicular line from P to the x-axis, we will get a right triangle. Then OP be the hypotenuse.
OP = √(-12)² + (-5)² OP = √(144 + 25) OP = √169 OP = 13 |
Opposite side = -5 Adjacent side = -12 Hypotenuse = 13 |
sin θ = Opposite side / Hypotenuse
sin θ = -5/13
cos θ = Adjacent side / Hypotenuse
cos θ = -12/13
tan θ = Opposite side / Adjacent side
tan θ = -5/-12
tan θ = 5/12
Problem 3 :
Solution :
By drawing a perpendicular line from P to the x-axis, we will get a right triangle. Then OP be the hypotenuse.
OP = √8² + (-15)² OP = √(64 + 225) OP = √289 OP = 17 |
Opposite side = -15 Adjacent side = 8 Hypotenuse = 17 |
sin θ = Opposite side / Hypotenuse
sin θ = -15/17
cos θ = Adjacent side / Hypotenuse
cos θ = 8/17
tan θ = Opposite side / Adjacent side
tan θ = -15/8
Problem 4 :
Solution :
OP = √ (1)² + (-1)² OP = √ (1 + 1) OP = √2 |
Opposite side = -1 Adjacent side = 1 Hypotenuse = √2 |
sin θ = Opposite side / Hypotenuse
sin θ = -1/√2
sin θ = -√2/2
cos θ = Adjacent side / Hypotenuse
cos θ = 1/√2
cos θ = √2/2
tan θ = Opposite side / Adjacent side
tan θ = -1/1
tan θ = -1
Determine the exact of sin θ, cos θ, and tan θ if the terminal arm of an angle in standard position passes through the given point.
Problem 5 :
P (-5, 12)
Solution :
Since the given point is in the form of (-x, y), it will be in second quadrant.
Accordingly ASTC, for the trigonometric ratios sin θ and cosec θ alone we have positive.
Finding hypotenuse : √(-5)² + (12)² √(25 + 144) √169 13 |
Opposite side = 12 Adjacent side = -5 Hypotenuse = 13 |
sin θ = Opposite side / Hypotenuse
sin θ = 12/13
cos θ = Adjacent side / Hypotenuse side
cos θ = -5/13
tan θ = Opposite side / Adjacent side
tan θ = -12/5
Problem 6 :
P (5, -3)
Solution :
Since the given point is in the form of (x, -y), it will be in fourth quadrant.
Accordingly ASTC, for the trigonometric ratios cos θ and sec θ alone we have positive.
Finding hypotenuse : √5² + (-3)² √(25 + 9) √34 |
Opposite side = -3 Adjacent side = 5 Hypotenuse = √34 |
sin θ = Opposite side / Hypotenuse side
sin θ = -3/√34
cos θ = Adjacent side / Hypotenuse side
cos θ = 5/√34
tan θ = Opposite side / Adjacent side
tan θ = -3/5
Problem 7 :
P (6, 3)
Solution :
Since the given point is in the form of (x, y), it will be in fourth quadrant.
Accordingly ASTC, for all trigonometric ratios we have to use positive sign.
Finding hypotenuse : √6² + (3)² √(36 + 9) √45 3√5 |
Opposite side = 3 Adjacent side = 6 Hypotenuse side = 3√5 |
sin θ = Opposite side / Hypotenuse
sin θ = 3√5
= 1/√5
sin θ = 1/√5
cos θ = Adjacent side / Hypotenuse
cos θ = 6/ 3√5
cos θ = 2/√5
tan θ = Opposite side / Adjacent side
tan θ = 3/6
tan θ = 1/2
Problem 8 :
P (-24, -10)
Solution :
Since the given point is in the form of (-x, -y), it will be in third quadrant.
Accordingly ASTC, for the trigonometric ratios tan θ and cot θ alone we have positive.
Finding hypotenuse : √(-24)² + (-10)² √(576 + 100) √676 26 |
Opposite side = -10 Adjacent side = -24 Hypotenuse = 26 |
sin θ = Opposite side / Hypotenuse
sin θ = -10/26
sin θ = -5/13
cos θ = Adjacent side / Hypotenuse
cos θ = -24/26
cos θ = -12/13
tan θ = Opposite side / Adjacent side
tan θ = -10/-24
tan θ = 5/12
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM