Sketch the angle in standard position

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Solutions
1. Consider an angle 5/4π rad
a) Sketch the angle in standard position
Solution

π

5 π/4
6 π/4
b) Determine 2 conterminal angles
Solution
I.
II.

5 π/4 +2 π =13 π/4
5 π/4 - 2 π=-3 π/4

c) Convert the angle to degree measures
Solution

5 π/4 * 180/ π =225°

2. A track moving at a rate of 90km/h and the diameter is 1.25m. Find the angular velocity of
Solution
The linear speed of wheel = 90km/h,
90km/h= (90*1000)/(3600) v=25m/s
radius of wheel path r = (1.25/2)=0.625
Angular Speed ω =v*r
=25*0.625

3. Find the exact values of the six trigonometric functions of the angle shown in the figure
Solution

When point (-1,4) i.e. (x=-1,y=4) is on the terminal side of an angle θ in standard position the
trigonometric functions are defined using radius(difference between the terminals) and the
origin:

=√ 17
1. Sin θ=y/r in this case sin θ= 4/√ 17

2. Cos θ=x/r in this case cos θ= -1/√ 17

3. Tan θ= y/x in this case tan θ= 4/-1 =-4

4. Cosec θ= r/y = √ 17/4

5. Sec θ=r/x= √ 17/-1
=-√ 17

6. Cot θ=x/y= -1/4
=-1/4

4. Given Tan θ= 7/2 and θ is an acute angle , find the other trigonometric
functions of θ

Solution
Opp=7
I.

Sin θ=opp/hypotenuse
Hypotenuse==√ (7)2+(2)2
=√ 53
=7/√ 53

II.

=2/ √ 53

5. Determine the reference angle θ’ of the angle θ=225 and sketch θ’ and θ in standard

position
Solution
Reference angle of 225°
255 lies in the third quadrant hence the formula is
=360°−ac
Ac=225°
=360-225

θ ‘=135°

θ ‘=135* π/180
Sketch

3π/4 (θ’)

π

5π/4(θ)
6 π/4

6. Determine the quadrant in which θ lies if Sec θ0

Solution
Sec θ=1/cos θ
And what is required 1/cos θ 0 and this will occur in quadrant 1&3
Hence they overlap in quadrant 3
7. Find two exact values of θ in degrees if cos θ=-√2/2

Solution
From special triangle
Cos θ= 1/√2

=45°

Hence cos245=1/2=2/4=√2/2

Therefore the angles lies between 2nd and 3rd quadrant since cos is negative
180-45=135°
180+45=225°
8. Use calculator to approximate 2 values of θ radians if θ =csc 1.030

Solution
csc θ = 1.03
csc θ=1/sin θ
sin θ = 1/1.03 = 0.971

θ = 1.33
also you can compute
π-1.33 = 1.8127

θ = 76.14°
or

θ= 103.86°
9. Find the remaining trigonometric functions of θ given that cos θ= =3/5 and sin θ >0

Solution
Cos θ=-3/5

Then θ is in quadrant, if x and y are positive numbers we will have
(-x,y) is a point in quadrant 2.
hence (r cos(θ), r sin(θ) )= (-x,y)
sin(θ) = y/r
cos(θ) = -x/r = -3/5
x = 3, r = 5
r2= x2+ y2
25 = 9 + y2
=4

sin(θ) = y/r = 4/5
cos(θ) = -x/r = -3/5
tan(θ) = sin(θ)/cos(θ) = -4/3
cot(θ) = 1/tan(θ) = -3/4
sec(θ) = 1/cos(θ) = -5/3
csc(θ) = 1/sin(θ) ...

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Anonymous
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