# EVALUATING TRIGONOMETRIC FUNCTIONS GIVEN A POINT ON THE TERMINAL SIDE

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 = √25OP = 5 Opposite side = 4Adjacent side = 3Hypotenuse = 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)²OP = √(144 + 25)OP = √169OP = 13 Opposite side = -5Adjacent side = -12Hypotenuse = 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)²OP = √(64 + 225)OP = √289OP = 17 Opposite side = -15Adjacent side = 8Hypotenuse = 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)²OP = √ (1 + 1)OP = √2 Opposite side = -1Adjacent side = 1Hypotenuse = √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)√16913 Opposite side = 12Adjacent side = -5Hypotenuse = 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 = -3Adjacent side = 5Hypotenuse = √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)√453√5 Opposite side = 3Adjacent side = 6Hypotenuse 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)√67626 Opposite side = -10Adjacent side = -24Hypotenuse = 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

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