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Trigonometry : Trigonometry

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6 Diagnostic Tests 155 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #8 : Arc Length

You know that the central angle of a sector is $\frac{3\pi}{2}$ and the sector area is $\frac{7\pi}{4}$ , what is the arc length? Round to two decimals.

Possible Answers:

l_A = 9.39

l_A = 2.51

l_A = 12.76

l_A = 14.67

Correct answer:

l_A = 12.76

Explanation:

The formula to find the area of a sector is $A = \frac{\theta}{2} r^2$ . The piece of information we are missing to solve for arc length is the radius, so we will use the formula for finding the area of a sector to solve for radius, allowing us to solve for the arc length.

$A = \frac{\theta}{2} r^2$

$\frac{7\pi}{4} = (\frac{3\pi}){2}(\frac{1}{2})r^2$

$\frac{7\pi}{4} = (\frac{3\pi}{4} r^2$

$\frac{7\pi}{3} = r^2$

$\sqrt{\frac{7\pi}{3}} = r$

2.707 = r

Now we can plug this into our formula to solve for arc length

$l_A = \theta r$

$l_A = \frac{3\pi}{2} (2.707)$

l_A = 12.76

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Example Question #591 : Trigonometry

Find all angles $\theta$ between $0^\circ$ and $360^\circ$ when $\cos \theta=-.3333$ .

Possible Answers:

$\theta=109.47^\circ$ and $250.53^\circ$

$\theta=70.53^\circ$ and $190.47^\circ$

$\theta=70.53^\circ$

$\theta=190.47^\circ$

$\theta=70.53^\circ$ and $250.53^\circ$

Correct answer:

$\theta=109.47^\circ$ and $250.53^\circ$

Explanation:

This problem relies on understanding reference angles and coterminal angles. A reference angle for an angle $\theta$ in standard position is the positive acute angle between the x axis and the terminal side of the angle $\theta$ . A table of reference angles for each quadrant is given below.

Screen shot 2020 07 30 at 11.05.57 am

Since $\cos \theta=-.3333$ is negative, solutions for $\theta$ will be in Quadrants II and III because these are the quadrants where cosine is negative.

Use inverse cosine and a calculator to find $\theta$ :

$\cos R = -.3333$

$\cos^-^1(.3333)=70.53^\circ = R$

In Quadrant II, we have $R=180^\circ-\theta$ , so $\theta=180^\circ-70.53^\circ=109.47^\circ$ .

In Quadrant III, $R=\theta-180^\circ$ , so $\theta=180^\circ+R=180^\circ+70.53^\circ=250.53^\circ$ .

Therefore $\theta=109.47^\circ$ and $250.53^\circ$ .

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Example Question #592 : Trigonometry

Find all angles $\theta$ between $0^\circ$ and $360^\circ$ when $\tan\theta=-1.3456$ .

Possible Answers:

$\theta=53.38^\circ$

$\theta=53.38^\circ$ and $306.62^\circ$

$\theta=126.62^\circ$ and $306.62^\circ$

$\theta=126.62^\circ$

$\theta=53.38^\circ$ and $126.62^\circ$

Correct answer:

$\theta=126.62^\circ$ and $306.62^\circ$

Explanation:

Screen shot 2020 07 30 at 11.05.57 am

Since $\tan\theta=-1.3456$ is negative, solutions for $\theta$ will be in Quadrants II and IV because these are the quadrants where tangent is negative. Use inverse tangent and a calculator to find $\theta$ :

$\tan R=-1.3456$

$\tan^-^1(1.3456)=53.38^\circ=R$

In Quadrant II, we have $R=180^\circ-\theta$ , so $\theta=180^\circ-53.38^\circ=126.62^\circ$ .

In Quadrant IV, $R=360^\circ-\theta$ , so $R=360^\circ-53.38^\circ=306.62^\circ$ .

Therefore $\theta=126.62^\circ$ and $306.62^\circ$ .

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Example Question #593 : Trigonometry

Find all angles $\theta$ when $\sin \theta=-.1919$ .

Possible Answers:

$\theta=11.06^\circ$ and $\theta=191.06^\circ$

$\theta=11.06^\circ+n360^\circ$ and $\theta=191.06^\circ+n360^\circ$

$\theta=191.06^\circ$ and $\theta=348.94^\circ$

$\theta=191.06^\circ+n360^\circ$ and $\theta=348.94^\circ+n360^\circ$

Correct answer:

$\theta=191.06^\circ+n360^\circ$ and $\theta=348.94^\circ+n360^\circ$

Explanation:

We can use reference angles, inverse trig, and a calculator to solve this problem. Below is a table of reference angles.

Screen shot 2020 07 30 at 11.05.57 am

We have $\sin \theta=-.1919$ so $\sin^-^1(.1919)=11.06^\circ = R$ . Next, think about where sine is negative, or reference the Function Signs column of the above table. Sine is negative in Quadrants III and IV.

In Quadrant III, $\theta=180^\circ+R=180^\circ+11.06^\circ=191.06^\circ$ .

In Quadrant IV, $\theta=306^\circ-R=360^\circ-11.06^\circ=348.94^\circ$ .

If this problem asked for values of $\theta$ between $0^\circ$ and $360^\circ$ , our work would be done, but this problem does not restrict the range, so we need to give all possible values of $\theta$ by generalizing our answers. To do this, we must understand that all angles that are coterminal to $191.06^\circ$ and $348.94^\circ$ will also be solutions. Coterminal angles add or subtract multiples of $360^\circ$ . To write this generally, we write:

$\theta=191.06^\circ+n360^\circ$ and $\theta=348.94^\circ+n360^\circ$ .

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Example Question #594 : Trigonometry

Find all angles $\theta$ when $\tan\theta=.5432$ .

Possible Answers:

$\theta=28.51^\circ+n360^\circ$ and $\theta=151.49^\circ+n360^\circ$

$\theta=28.51^\circ$ and $\theta=208.51^\circ$

$\theta=28.51^\circ+n360^\circ$ and $\theta=208.51^\circ+n360^\circ$

$\theta=28.51^\circ$ and $\theta=151.49^\circ$

Correct answer:

$\theta=28.51^\circ+n360^\circ$ and $\theta=208.51^\circ+n360^\circ$

Explanation:

We can use reference angles, inverse trig, and a calculator to solve this problem. Below is a table of reference angles.

Screen shot 2020 07 30 at 11.05.57 am

We have $\tan R=.5432$ , so $\tan ^-^1(.5432)=28.51^\circ=R$ . Next, think about where tangent is positive, or reference the Function Signs column of the above table. Tangent is positive in Quadrants I and III.

In Quadrant I, $\theta=R=28.51^\circ$ .

In Quadrant III, $\theta=180^\circ+R=180^\circ+28.51^\circ=208.51^\circ$ .

If this problem asked for values of $\theta$ between $0^\circ$ and $360^\circ$ , our work would be done, but this problem does not restrict the range, so we need to give all possible values of $\theta$ by generalizing our answers. To do this, we must understand that all angles that are coterminal to $28.51^\circ$ and $208.51^\circ$ will also be solutions. Coterminal angles add or subtract multiples of $360^\circ$ . To write this generally, we write:

$\theta=28.51^\circ+n360^\circ$ and $\theta=208.51^\circ+n360^\circ$

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Example Question #595 : Trigonometry

Find all positive values of $\theta$ less than $360^\circ$ for which $\cos\theta=-\frac{1}{2}$ .

Possible Answers:

$\theta=120^\circ$ and $240^\circ$

$\theta=60^\circ$ and $240^\circ$

$\theta=240^\circ$

$\theta=60^\circ$

$\theta=120^\circ$

Correct answer:

$\theta=120^\circ$ and $240^\circ$

Explanation:

At first glance, you may think that this problem has infinite answers, since there would be infinitely many negative coterminal angles that could satisfy this; however, notice that the question asks only for positive values of $\theta$ . In other words, this question is simply asking for values of $\theta$ between $0^\circ$ and $360^\circ$ that satisfy this equation.

First, let's think about where the cosine function is negative. Per the chart below, it will be in Quadrants II and III.

Screen shot 2020 07 30 at 10.25.02 am

The reference angle for each angle solution will have its cosine equal to $+\frac{1}{2}$ and is $60^\circ$ . Consult the chart of reference angles below for Quadrants II and III:

Screen shot 2020 07 30 at 11.05.57 am

QII: $\theta=180^\circ-R=180^\circ-60^\circ=120^\circ$

QIII: $\theta=R+180^\circ=60^\circ+180^\circ=240^\circ$

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Example Question #596 : Trigonometry

Find all angles $\theta$ between $0^\circ$ and $360^\circ$ when $\cos\theta=.8999$ .

Possible Answers:

$\theta=334.14^\circ$

$\theta=25.86^\circ$

$\theta=25.86^\circ$ and $154.14^\circ$

$\theta=154.14^\circ$

$\theta=25.86^\circ$ and $334.14^\circ$

Correct answer:

$\theta=25.86^\circ$ and $334.14^\circ$

Explanation:

Screen shot 2020 07 30 at 11.05.57 am

Since $\cos\theta=.8999$ is positive, solutions for $\theta$ will be in Quadrants I and IV because these are the quadrants where cosine is positive. Use inverse cosine and a calculator to find $\theta$ :

$\cos\theta=.8999$

$\cos^-^1=(.8999)=25.86^\circ=R$

In Quadrant I, we have $R=\theta$ , so $\theta=25.86^\circ$ .

In Quadrant IV, $R=360^\circ-\theta$ , so $R=360^\circ-25.86^\circ=334.14^\circ$ .

Therefore $\theta=25.86^\circ$ and $334.14^\circ$ .

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Example Question #597 : Trigonometry

Find all angles $\theta$ between $0^\circ$ and $360^\circ$ when $\sin \theta=.6478$ .

Possible Answers:

$\theta=$ $139.62^\circ$ and $319.62^\circ$

$\theta = 40.38^\circ$ and $139.62^\circ$

$\theta = 40.38^\circ$ and $319.62^\circ$

$\theta=$ $130.38^\circ$ and $139.62^\circ$

$\theta = 40.38^\circ$ and $130.38^\circ$

Correct answer:

$\theta = 40.38^\circ$ and $139.62^\circ$

Explanation:

Screen shot 2020 07 30 at 11.05.57 am

Since $\sin \theta=.6478$ is positive, solutions for $\theta$ will be in Quadrants I and II because these are the quadrants where sine is positive. Use inverse sine and a calculator to find $\theta$ :

$\sin R=.6478$

$\sin^-^1(.6478)=40.38^\circ$

In Quadrant I, we have $R=\theta$ , so $\theta=40.38^\circ$ .

In Quadrant II, $R=180^\circ-\theta$ , so $R=180^\circ-40.38^\circ=139.62^\circ$ .

Therefore $\theta = 40.38^\circ$ and $139.62^\circ$ .

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Example Question #1 : Practical Applications

Select the answer that correctly matches the following air navigation terms to their definitions.

Possible Answers:

The heading of an airplane is the direction in which the airplane is pointed. The heading is measured clockwise from the north and expressed in degrees.

The airspeed is the speed of the airplane in still air.

The course of an airplane is the direction in which it moves relative to the ground. The course is measured clockwise from the north.

The groundspeed is the speed of the airplane relative to the ground.

The drift angle is the positive difference between the heading and the course.

The heading of an airplane is the direction in which the airplane is pointed. The heading is measured clockwise from the north and expressed in degrees.

The groundspeed is the speed of the airplane in still air.

The course of an airplane is the direction in which it moves relative to the ground. The course is measured clockwise from the north.

The airspeed is the speed of the airplane relative to the ground.

The drift angle is the positive difference between the heading and the course.

The course of an airplane is the direction in which the airplane is pointed. The heading is measured clockwise from the north and expressed in degrees.

The groundspeed is the speed of the airplane in still air.

The heading of an airplane is the direction in which it moves relative to the ground. The course is measured clockwise from the north.

The airspeed is the speed of the airplane relative to the ground.

The drift angle is the positive difference between the heading and the course.

The course of an airplane is the direction in which the airplane is pointed. The heading is measured clockwise from the north and expressed in degrees.

The airspeed is the speed of the airplane in still air.

The heading of an airplane is the direction in which it moves relative to the ground. The course is measured clockwise from the north.

The groundspeed is the speed of the airplane relative to the ground.

The drift angle is the positive difference between the heading and the course.

Correct answer:

The heading of an airplane is the direction in which the airplane is pointed. The heading is measured clockwise from the north and expressed in degrees.

The airspeed is the speed of the airplane in still air.

The course of an airplane is the direction in which it moves relative to the ground. The course is measured clockwise from the north.

The groundspeed is the speed of the airplane relative to the ground.

The drift angle is the positive difference between the heading and the course.

Explanation:

The heading of an airplane is the direction in which the airplane is pointed. The heading is measured clockwise from the north and expressed in degrees.

The airspeed is the speed of the airplane in still air.

The course of an airplane is the direction in which it moves relative to the ground. The course is measured clockwise from the north.

The groundspeed is the speed of the airplane relative to the ground.

The drift angle is the positive difference between the heading and the course.

You may use vectors to represent airspeed and heading, direction and speed of wind, or groundspeed and course. The groundspeed vector is the resultant of the airspeed vector and the wind vector.

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Example Question #1 : Inclined Planes And Air Navigation

In the following diagram, a blue box sits on an inclined plane. The box has weight W and exerts force F_a against the inclined plane and force F_d down the inclined plane. Which of the following correctly relates these vectors together?

Screen shot 2020 08 03 at 5.16.21 pm

Possible Answers:

F_a-W=F_d

F_d-W=F_a

W+F_a=F_d

W+F_d=F_a

F_d+F_a=W

Correct answer:

F_d+F_a=W

Explanation:

The correct answer is F_d+F_a=W because F_d and F_a are component vectors for the weight .

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Trigonometry : Trigonometry

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #8 : Arc Length

Example Question #591 : Trigonometry

Example Question #592 : Trigonometry

Example Question #593 : Trigonometry

Example Question #594 : Trigonometry

Example Question #595 : Trigonometry

Example Question #596 : Trigonometry

Example Question #597 : Trigonometry

Example Question #1 : Practical Applications

Example Question #1 : Inclined Planes And Air Navigation

All Trigonometry Resources

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