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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 #251 : Trigonometry

Solve for :

$0 = 4 \cos ^2 x - 21 \cos x + 5$

Possible Answers:

14.47^o, 165.52^o

x = 75.52^o, 255.52 ^o

x = 75.52^o, 284.48^o

x = 75.52^o, 104.48^o

14.47^o, 345.52^o

Correct answer:

x = 75.52^o, 284.48^o

Explanation:

Solve using the quadratic formula:

$\cos x = \frac{21 \pm \sqrt{441 - 4(4)(5)}}{2(4) } = \frac{ 21 \pm \sqrt{361}}{8} = \frac{21 \pm 19}{8}$

$\cos x = 5, \frac{1}{4}$

5 is outside the range for cosine, so the only solution that works is $\cos x = \frac{1}{4}$ :

$x = \cos ^{-1} (\frac{1}{4}) \approx 75.52 ^o$ according to a calculator

The other angle with a cosine of $\frac{1}{4}$ is 360^o - 75.52^o = 284.48^o

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Example Question #252 : Trigonometry

Solve for : $0 = 3 \cos ^2 x + 8 \cos x + 4$

Possible Answers:

48.19^o, 228.19^o

131.81^o, 311.81^o

131.81^o, 228.19^o

48.19^o, 311.81^o

48.19^o, 131.81^o

Correct answer:

131.81^o, 228.19^o

Explanation:

Use the quadratic formula:

$\cos x = \frac{ -8 \pm \sqrt{64 - 4(3)(4)}}{2(3)} = \frac{-8 \pm \sqrt{16}}{6 } = \frac{-8\pm4}{6}$

$\cos x = -2 \enspace or - \frac{2}{3}$

-2 is outside the range of cosine, so the answer has to come from $- \frac{2}{3}$ :

$\cos x = - \frac{2}{3}$

$x = \cos^{-1} (-\frac{2}{3} ) \approx 131.81^o$ according to a calculator

The other angle with a cosine of $-\frac{2}{3}$ is 360^o - 131.81^o = 228.19^o

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Example Question #253 : Trigonometry

Solve the equation

2cos^2x+sinx=1

for $0\leq x\leq2\pi$ .

Possible Answers:

$\frac{\pi}{2},\frac{\pi}{6}$

$\frac{\pi}{2},\frac{7\pi}{6}, \frac{11\pi}{6}$

$\frac{\pi}{2}$

$\frac{\pi}{2},\frac{\pi}{6}, \frac{2\pi}{3}$

$\frac{7\pi}{6},\frac{11\pi}{6}$

Correct answer:

$\frac{\pi}{2},\frac{7\pi}{6}, \frac{11\pi}{6}$

Explanation:

First of all, we can use the Pythagorean identity sin^2x+cos^2x=1 to rewrite the given equation in terms of sinx .

2cos^2x+sinx=1

$\Rightarrow 2(1-sin^2x)+sinx=1$

$\Rightarrow 2-2sin^2x+sinx=1$

$\Rightarrow -2sin^2x+sinx+1=0$

$\Rightarrow 2sin^2x-sinx-1=0$

This is a quadratic equation in terms of sinx ; hence, we can use the quadratic formula to solve this equation for sinx .

$sinx=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

where a=2, b=-1, c=-1 .

$\Rightarrow sinx=\frac{1\pm \sqrt{(-1)^2-4(2)(-1)}}{2(2)}$

$\Rightarrow sinx=\frac{1\pm3}{4}$

$\Rightarrow sinx=-\frac{1}{2}, sinx=1$ .

Now, sinx=1 when $x=\frac{\pi}{2}$ , and $sinx=-\frac{1}{2}$ when $x=\frac{7\pi}{6}$ or $\frac{11\pi}{6}$ .

Hence, the solutions to the original equation 2cos^2x+sinx=1 are

$\frac{\pi}{2},\frac{7\pi}{6}, \frac{11\pi}{6}$

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Example Question #254 : Trigonometry

In the interval $0\leq x\leq\pi$ , what values of x satisfy the following equation?

$1-\sin x=\cos^2x-\frac{1}{4}$

Possible Answers:

$x=0,\pi$

$x=\frac{\pi}{3},\frac{2\pi}{3}$

$x=\frac{\pi}{6},\frac{5\pi}{6}$

$x=\frac{\pi}{4},\frac{7\pi}{12}$

Correct answer:

$x=\frac{\pi}{6},\frac{5\pi}{6}$

Explanation:

We start by rewriting the $\cos^2x$ term on the right hand side in terms of $\sin^2x$ .

$\cos^2x=1-\sin^2x$

$1-\sin x=1-\sin^2x-\frac{1}{4}$

We then move everything to the left hand side of the equation and cancel.

$1-\sin x-1+\sin^2x+\frac{1}{4}=0$

$\sin^2x-\sin x+\frac{1}{4}=0$

Apply the quadratic formula:

$\begin{align*} \sin x&=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(\frac{1}{4})}}{2(1)}\\ &=\frac{1\pm\sqrt{1-1}}{2}\\ &=\frac{1}{2} \end{align*}$

So $\sin x=\frac{1}{2}$ . Using the unit circle, the two values of that yield this are $x=\frac{\pi}{6}$ and $x=\frac{5\pi}{6}$ .

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Example Question #1 : Trigonometric Applications

Given a right triangle where $\sin\theta = \frac{3}{5}$ , find the missing side.

Possible Answers:

${}\sqrt{8}$

$\sqrt{41}$

$\sqrt{34}$

Correct answer:

Explanation:

Since the triangle in question is a right triangle we can use the Pythagorean Theorem. First, we must determine which sides we are given. Since the function we are given is sine, we know that we are given the opposite side and the hypotenuse. Therefore, setting up the equation:

$a^{2}+b^{2}=c^{2}$

Where, and are given.

$3^{2} + b^{2} = 5^{2}$

Solving the above equation:
$b = \sqrt{16} = 4$

We toss out the negative solution since the length of a side must be positive.

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Example Question #2 : Trigonometric Applications

Given a right triangle where $\tan\theta = \frac{17}{18}$ , find the missing side.

Possible Answers:

24.76

24.67

5.92

17.01

5.29

Correct answer:

24.76

Explanation:

Since the triangle in question is a right triangle we can use the Pythagorean Theorem. First, we must determine which sides we are give. Since the function we are given is tangent, we know that we are given the opposite and adjacent sides. Therefore, setting up the equation:

$a^{2}+b^{2}=c^{2}$

Where, and are given.

$17^{2}+18^{2} = c^{2}$

Solving the above equation:

$c = \sqrt{613} = 24.76$

We toss out the negative solution since the length of a side must be positive.

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Example Question #3 : Trigonometric Applications

Given a right triangle where $\cos\theta = \frac{104}{109}$ , find the missing side.

Possible Answers:

107

32.36

150.66

32.63

150.56

Correct answer:

32.63

Explanation:

Since the triangle in question is a right triangle we can use the Pythagorean Theorem. First, we must determine which sides we are given. Since the function we are given is cosine, we know that we are given the adjacent side and hypotenuse. Therefore, setting up the equation:

$a^{2}+b^{2}=c^{2}$

Where, and are given.

$104^{2} + b^{2} = 109^{2}$

Solving the above equation:

$b=\sqrt{1,065}=32.63$

We toss out the negative solution since the length of a side must be positive.

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Example Question #4 : Trigonometric Applications

Given the accompanying triangle where b=6 and $\theta = 17^{\circ}$ , determine the length of the hypotenuse.

Right_triangle

Possible Answers:

1.75

5.74

20.52

1.83

19.63

Correct answer:

20.52

Explanation:

We are given the opposite side, with respect to the angle, along with the angle. Therefore, we utilize the sine function to determine the length of the hypotenuse:

$\sin\theta = \frac{opposite}{hypotenuse}$

Substituting the given values:

$\sin 17^{\circ} = \frac{6}{h}$

Cross multiplying:

$h\sin 17^{\circ}=6$

Solving for :

$h = \frac{6}{\sin 17^{\circ}} = 20.52$

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Example Question #1 : Triangles

Given the accompanying right triangle where b=7 and c=13 , determine the measure of $\theta$ to the nearest degree.

Right_triangle

Possible Answers:

$67^{\circ}$

$33^{\circ}$

$30^{\circ}$

$68^{\circ}$

$36^{\circ}$

Correct answer:

$33^{\circ}$

Explanation:

We are given two sides of the right triangle, namely the hypotenuse and the opposite side of the angle. Therefore, we simply use the sine function to determine the angle:
$\sin\theta = \frac{opposite}{hypotenuse} =\frac{7}{13}$

In order to isolate the angle we must apply the inverse sine function to both sides:

$\theta = sin^{-1}\frac{7}{13} = 32.57^{\circ} \approx 33^{\circ}$

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Example Question #6 : Trigonometric Applications

Given the accompanying right triangle where a=35 and c=70 , determine the measure of $\theta$ to the nearest degree.

Right_triangle

Possible Answers:

$45^{\circ}$

$62^{\circ}$

$37^{\circ}$

$60^{\circ}$

$30^{\circ}$

Correct answer:

$60^{\circ}$

Explanation:

We are given two sides of the right triangle, namely the hypotenuse and the adjacent side of the angle. Therefore, we simply use the cosine function to determine the angle:

$\cos\theta = \frac{adjacent}{hypotenuse}=\frac{35}{70}$

In order to isolate the angle we must apply the inverse cosine function to both sides:

$\theta = \cos^{-1}\frac{35}{70}=\cos^{-1}\frac{1}{2}=60^{\circ}$

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Trigonometry : Trigonometry

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #251 : Trigonometry

Example Question #252 : Trigonometry

Example Question #253 : Trigonometry

Example Question #254 : Trigonometry

Example Question #1 : Trigonometric Applications

Example Question #2 : Trigonometric Applications

Example Question #3 : Trigonometric Applications

Example Question #4 : Trigonometric Applications

Example Question #1 : Triangles

Example Question #6 : Trigonometric Applications

All Trigonometry Resources

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