Precalculus II work

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

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

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

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 =

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.

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.

Sec θ = - 5sqrt(19)/19

Option (a) is correct.

Question 6 In the triangle below, c= 173 cm. Solve the triangle for sides a and b, rounding to two decimal places.

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:

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

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?

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

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 -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

When we compare general equation and y = 6 sin(x)

We will get the value of b = 1/3

Then,

Period = 2pi/b

= 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.

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

Solution:

Amplitude = 6, period = 2pi/ (pi/20) = 40

Option b have fulfill this criteria Amplitude = 6 and period = 40

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)

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)

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****