Download Chapter 6 test Question 1 Find the reference angle. θ = 240° a. 30° b

Chapter 6 test
Question 1 Find the reference angle. θ = 240°
a. 30° b. 40° c. 50° d. 60°
Solution:
We know the 360 degrees is the total circle angle. The reference
angle is referenced based upon this.
So, the angle is -240 you would subtract that from 360.
=360-240 = 60 degree
Chapter 6 test
Question 2 Find the reference angle. θ = 105°
a. 105° b. 75° c. 15° d. 195°
Solution:
First, draw 105 degrees. You should end up in the second
quadrant.
Now draw a line (terminal line) diagonally from the origin. Your
reference angle is the angle that is created between the x-axis
and the terminal line. So, to get the reference angle, you must
subtract 105 degrees from 180 degrees.
= 180 - 105 = 75 Degree
Chapter 6 test
Question 3 Find the reference angle. Θ=-7π/4
a. π/4 b. 3π/4 c. – π/4 d. -3π/4
Solution:
A reference angle is just another way to write the same angle.
Because a full circle is 2π, we just add 2π to -7π/4.
8π/4 - 7π/4 = π/4 radian
Chapter 6 test
Solution:
Given tan theta = -sqrt(55)/5
We know from trigonometric formula
Sec^2 θ= 1 + tan^2 θ
Plug the value of tan theta
Sec^2 θ = 1+ (-sqrt(55)/5)
= 1+ 55/25 = 80/25
Take Square root both sides.
Sec θ =
Chapter 6 test
Sec θ = 4sqrt(5)/5 or cos θ = 5/ 4sqrt(5)
We know,
Tan θ = sin θ/cos θ
Sin θ = tan θ * cos θ
Plug the value of tan theta and cos theta
Sin θ = (-sqrt(55)/5) *(5/ 4sqrt(5))
= -sqrt(11)/4 The angle lies in the IV quadrant that is why sin will be negative.
So the Sin theta = - (-sqrt(11)/4) = sqrt(11)/4
Option (d) is correct.
Chapter 6 test
Solution:
Given tan theta = -sqrt(114)/19
We know from trigonometric formula
Sec^2 θ = 1 + tan^2 θ
Plug the value of tan theta
Sec^2 θ = 1+ (-sqrt(114)/19)^2 = 1+ 114/361 = 475/361
Take Square root both sides.
Sec θ = 5sqrt(19)/19 , the angle lie in the II quadrant that is
why sec theta will be negative.
Chapter 6 test
Sec θ = - 5sqrt(19)/19
Option (a) is correct.
c= 173 cm. Solve the triangle for
sides a and b, rounding to two decimal places.
Question 6 In the triangle below,
a. a = 222.61 cm , b = 274.90 cm b. a = 274.90 cm, b = 222.61 cm
b. a = 134.45 cm, b = 108.87 cm d. a = 108.87 cm, b = 134.45 cm
Solution:
Chapter 6 test
Apply the formula in above triangle.
Sin 39 = adjacent/ hypotenuse
Adjacent = sin 39 * hypotenuse
Plug the value of hypotenuse (c = 173 cm) and we know the
value of sin 39 = 0.6293
Adjacent (a) or (leg 1) = 0.63 * 173 = 108.87 centimeter
Now we have two value a = 108.87 cm and c = 173 cm.
Apply the Pythagorean Theorem and find the value f b.
Adjacent (c) or (leg 2) = Sqrt(c^2 – b^2)
= Sqrt (173^2 -108.87^2)
= 134.45 cm
Chapter 6 test
Question 7 Find the measure of angle θ to the nearest tenth of a degree.
10 yd
20 yd
a. 63.4° b. 64.3° c. 64.7° d. 65.1°
Solution:
Apply the formula in above triangle.
Tan theta = 20 yd/10 yd = 2
Theta = Arc tan (2)
Theta = 63.43 degree
Question 8 A person is standing 250 feet from a building. The angle to the
top of the building relative to the horizon is 63.1°. How tall is the
building?
Chapter 6 test
a. 491.4 ft b. 492.8 ft c. 494.0 ft d. 494.6 ft
Solution:
From the question the diagram would be like this.
Now, we apply the formula in a triangle ABC.
Tan 63.1 = BC/AC
Chapter 6 test
Plug the value of AC = 250 feet.
Height of the building (BC) = 250 feet * Tan 63.1
= 492.8 feet
Question 9
y = - 8 sin(
x)
State the amplitude.
a. 8 b. –8 c. 5 d. –5
Solution:
Look at the general equation:
y = a sin (b (x + c)) + d
Where a is the amplitude, b is the frequency, c is the
horizontal shift and d is the vertical shift.
In a trigonometric function, the period is equal to 2π / b
When we compare both the equation we will get the amplitude a
Chapter 6 test
-8
Question 10:
y = 6 sin
( x)
State the period.
a. π/3 b.2π/3 c. 3π d. 6π
Solution:
We know that
Period = 2π / b
( x)
When we compare general equation and y = 6 sin
We will get the value of b = 1/3
Then,
Period = 2pi/b
Chapter 6 test
= 2pi/(1/3)
= 6pi
Solution:
In above graph Amplitude = -3 and Period = pi/4
So, pi/4 = 2pi/B, B = 8
W e have the equation in the form y = Asin(Bx)
Plug the values.
Equation (y) = =-3sin(8x) Option a is correct answer.
Chapter 6 test
Solution:
When we compare above equation to general equation then we
will have the amplitude 180
Solution:
As we know Period = 2pi/b, When we compare above equation to
general equation then we will have the b = pi/9.
Period = 2pi/ (pi/9)
= 18
Chapter 6 test
Solution:
Amplitude = 6, period = 2pi/ (pi/20) = 40
Option b have fulfill this criteria Amplitude = 6 and period =
40
Chapter 6 test
Solution:
Answer of this graph equation y = 5csc (2x) will be option d.
Solution:
We need to find exact value of Arcsin (- sqrt(2)/2)
We know from properties of inverse.
Arccos (-x) = pi- Arccos (x)
Chapter 6 test
We will apply this property here.
Arcsin (- sqrt(2)/2) = pi – Arccos(sqrt(2)/2)
Now we got the standard value of Arccos(sqrt(2)/2) = pi/4
So,
Arcsin (- sqrt(2)/2) = pi – pi/4 = 3pi/4
Solution:
We need to find exact value of Arcsin (-1).
We know from properties of inverse.
Arc sin (-x) = -Arc sin (x)
Chapter 6 test
We will apply this property here.
Arcsin (-1) = - Arcsin (1)
Now we got the standard value of Arcsin (1) = pi/2
So
- Arcsin (1) = - pi/2
***End of the solution****