SKETCH AN ANGLE IN STANDARD POSITION IF TERMINAL ARM PASSES THROUGH  A POINT

Example 1 :

Sketch an angle in standard position so that the terminal arm passes through each point.

(a) (2, 6)

Solution : 

First let us plot the point (2, 6) in the graph paper.

(b) (-4, 2)

(c) (-5, -2) 

(d) (-1, 0)

Example 2 :

Determine the exact values of the sine, cosine, and tangent ratios for each angle

Solution :

By drawing a perpendicular line from A, we get a triangle OAB.

In triangle OAB,

OA  =  Hypotenuse side

OB  =  Adjacent side and AB = Opposite side

Since the terminal arm lies in first quadrant, we have to take positive sign for sin 60.

(ii)  

By drawing a perpendicular line from A, we get a triangle OAB.

In triangle OAB,

<BOA  =  180 +  θ  

180 +  θ  = 225

 θ  =  225 - 180  =  45

Since the terminal arm lies in the 3rd quadrant, we have to use positive sign for tan and cot only.

(iii)

By drawing a perpendicular line from A, we get a triangle OAB.

In triangle OAB,

<AOB  =  180 - θ  

<AOB  =  180 - 150  

<AOB  =  30

(iv)

Example 3 :

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.

Solution :

Since the terminal arm lies in 3rd quadrant, we have to take positive signs for all trigonometric ratios.

(ii)

Since the terminal arm lies in 3rd quadrant, we have to take positive sign only for tan and cot.

(iii)

Since the terminal arm lies in 4th quadrant, we have to take positive sign only for cos and sec.

Example 4 :

a)  Plot the given point 𝑃(−4,−5). What quadrant does this point lie in?

b) Construct the corresponding angle in standard position.

c) Drop a perpendicular to the 𝑥-axis creating a right triangle. Which value represents the adjacent side? Which value represents the opposite side?

d)  How can you determine the exact length of the hypotenuse?

e)  State the cosine, sine and tangent ratios associated with the angle

f) What determines the sign of the ratio? Explain your reasoning.

Solution :

a) Since the given point is in the form of (-x, -y), it will lie in the third quadrant.

b)

c) 

• The side which is opposite to angle measure is opposite side.
• The side which is opposite to angle measure 90 degree is hypotenuse.
• The remaining side is adjacent side.

d)

Horizontal length = adjacent side = 4

Vertical lenght = Opposite side = 5

Hypotenuse = √4² + 5²

= √16 + 25

= √41

So, the required hypotenuse is √41

e) sin θ  = Opposite side/hypotenuse

cos θ  = Adjacent side/hypotenuse

tan θ  = Opposite side/adjacent side

f)

Using ASTC, for the trigonometric ratios tan and cot we have positive sign.

sin θ  = -4/√41

cos θ  = -5/√41

tan θ  = -4/(-5) ==> 4/5

Example 5 :

The point P(−3, 4) lies on the terminal arm of an angle, θ, in standard position. Determine the exact trigonometric ratio for cos θ, sin θ and tan θ.

Solution :

Opposite side = 4

Adjacent side = -3

Hypotenuse = √4² + (-3)²

= √16 + 9

= √25

= 5

Using ASTC, for the trigonometric ratios which lies in the second quadrant sin and cosec we will have positive sign.

sin θ = 4/5

cos θ = -3/5

tan θ = -4/3

Example 6 :

The point P(−8, −15) lies on the terminal arm of an angle, θ, in standard position. Determine the exact trigonometric ratio for cos θ, sin θ and tan θ.

Solution :

Opposite side = -15

Adjacent side = -8

Hypotenuse = √(-15)² + (-8)²

= √(225 + 64)

= √289

= 17

Using ASTC, for the trigonometric ratios which lies in the third quadrant tan and cot we will have positive sign.

sin θ = -15/17

cos θ = -8/17

tan θ = 15/8

Example 7 :

Suppose θ is an angle in standard position with terminal arm in quadrant III and cos θ = −3/4 . What are the exact values of sin θ and tan θ?

Solution :

cos θ = Adjacent side / hyopotenuse

cos θ = −3/4

Opposite side = √4² - (-3)²

= √(16 - 9)

= √7

Using ASTC, for the trigonometric ratios which lies in the third quadrant tan and cot we will have positive sign.

sin θ = Opposite side/hypotenuse

sin θ = -√7/4

tan θ = opposite side / adjancent side

tan θ = √7/3

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