THE GIVEN POINT LIES ON THE TERMINAL SIDE EVALUATING TRIG FUNCTIONS

Use the given point on the terminal side of angle θ to find the value of the trigonometric function indicated.

Problem 1 :

tan θ ; (8, 18)

Solution :

Since the terminal side lies in the first quadrant, all trigonometric ratios will be positive.

 tan θ = opposite/adjacent

= 18/8

tan θ  = 9/4

Problem 2 :

sec θ ; (√5, -2)

Solution :

Since the terminal side lies in the fourth quadrant, for cosine θ and its reciprocal sec θ are positive.

sec θ = hypotenuse/adjacent

Let 'r' be the hypotenuse.

By using Pythagorean theorem.

r² = x² + y²

r² = (√5)² + (-2)²

r² = 5 + 4

r² = 9

r = 3

sec θ = 3/√5

Problem 3 :

cos θ ; (√19, 9)

Solution :

Since the terminal side lies in the first quadrant, all trigonometric ratios will be positive.

cos θ = adjacent/hypotenuse

Let 'r' be the hypotenuse.

By using Pythagorean theorem.

r² = x² + y²

r² = (√19)² + (9)²

r² = 19 + 81

r² = 100

r = 10

cos θ = √19/10

Problem 4 :

sin θ ; (3, -4)

Solution :

Since the terminal side lies in the fourth quadrant, for cosine θ and its reciprocal sec θ are positive.

sin θ = opposite/hypotenuse

Let 'r' be the hypotenuse.

By using Pythagorean theorem.

r² = x² + y²

r² = 3² + (-4)²

r² = 9 + 16

r² = 25

r = 5

sin θ = -4/5

Problem 5 :

sin θ ; (-6, 4)

Solution :

Since the terminal side lies in the second quadrant, for sin θ and its reciprocal cosec θ are positive.

sin θ = opposite/hypotenuse

Let 'r' be the hypotenuse.

By using Pythagorean theorem.

r² = x² + y²

r² = (-6)² + (4)²

r² = 36 + 16

r² = 52

r = √52

= √(13 × 4)

= 2√13

sin θ = 4/2√13

sin θ = 2/√13 

Problem 6 :

cos θ ; (2, -2)

Solution :

Since the terminal side lies in the fourth quadrant, for cosine θ and its reciprocal sec θ are positive.

cos θ = adjacent/hypotenuse

Let 'r' be the hypotenuse.

By using Pythagorean theorem.

r² = x² + y²

r² = 2² + (-2)²

r² = 4 + 4

r² = 8

r = √8

cos θ = 2/√8

= 2/√(4 × 2)

= 2/2√2

cos θ = 1/√2

Problem 7 :

cot θ ; (-7, √15)

Solution :

Since the terminal side lies in the second quadrant, for sin θ and its reciprocal cosec θ are positive.

cot θ = adjacent/opposite

= √15/-7

Problem 8 :

cos θ ; (-5, -12)

Solution :

Since the terminal side lies in the third quadrant, for tan θ and its reciprocal cot θ are positive.

cos θ = adjacent/hypotenuse

Let 'r' be the hypotenuse.

By using Pythagorean theorem.

r² = x² + y²

r² = (-5)² + (-12)²

r² = 25 + 144

r² = 169

r = 13

cos θ = -5/13

Problem 9 :

sin θ ; (-√7, 3)

Solution :

Since the terminal side lies in the second quadrant, for sin θ and its reciprocal cosec θ are positive.

sin θ = opposite/hypotenuse

Let 'r' be the hypotenuse.

By using Pythagorean theorem.

r² = x² + y²

r² = (-√7)² + (3)²

r² = 7 + 9

r² = 16

r = 4

sin θ = 3/4

Problem 10 :

tan θ ; (-11, -2)

Solution :

Since the terminal side lies in the third quadrant, for tan θ and its reciprocal cot θ are positive.

tan θ = opposite/adjacent

= -2/-11

tan θ  = 2/11