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SAT II Math I : Trigonometry

Study concepts, example questions & explanations for SAT II Math I

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Example Questions

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Example Question #1 : Secant, Cosecant, Cotangent

Find the value of the trigonometric function in fraction form for triangle ABC .

Triangle

What is the secant of $\angle A$ ?

Possible Answers:

7/25

25/7

24/25

24/7

$\frac{25}{24}$

Correct answer:

$\frac{25}{24}$

Explanation:

The value of the secant of an angle is the value of the hypotenuse over the adjacent.

Therefore:

$sec \angle A = \frac{hypotenuse}{adjacent} = \frac{25}{24}$

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Example Question #21 : Trigonometry

Which of the following is the equivalent to $\frac{1}{\csc\theta}$ ?

Possible Answers:

$\sec\theta$

$\cot\theta$

$\tan\theta\sin\theta$

$\sin\theta$

$\cos\theta$

Correct answer:

$\sin\theta$

Explanation:

Since $\csc\theta=\frac{1}{\sin\theta}$ :

$\frac{1}{\csc\theta}=\frac{1}{\frac{1}{\sin\theta}}=1*\frac{\sin\theta}{1}=\sin\theta$

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Example Question #21 : Trigonometry

Soh_cah_toa

For the above triangle, what is $\sec (\theta)$ if o = 8 , a = 15 and h = 17 ?

Possible Answers:

2.13

{}1.13

0.53

0.88

0.47

Correct answer:

{}1.13

Explanation:

Secant is the reciprocal of cosine.

$\sec \left ( \theta\right ) = \frac{1}{\cos \left ( \theta\right )}$

It's formula is:

$\sec(\theta) = \frac{\textup{hypotenuse}}{\textup{adjacent}}$

Substituting the values from the problem we get,

$\sec(\theta) = \frac{17}{15} = 1.13$

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Example Question #1 : Sec, Csc, Ctan

Soh_cah_toa

For the above triangle, what is $\cot (\theta)$ if o = 18 , a = 24 and h = 30 ?

Possible Answers:

0.75

1.33

1.67

0.60

1.25

Correct answer:

1.33

Explanation:

Cotangent is the reciprocal of tangent.

$\cot \left ( \theta\right ) = \frac{1}{\tan \left ( \theta\right )}$

It's formula is:

$\cot(\theta) = \frac{\textup{adjacent}}{\textup{opposite}}$

Substituting the values from the problem we get,

$\cot(\theta) = \frac{24}{18} = 1.33$

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Example Question #25 : Trigonometry

Determine the value of $cot(\frac{\pi}{4})$ .

Possible Answers:

$\frac{\sqrt2}{2}$

$\infty$

$\frac{1}{2}$

Correct answer:

Explanation:

Rewrite $cot(\frac{\pi}{4})$ in terms of sine and cosine.

$cot(\frac{\pi}{4})= \frac{cos(\frac{\pi}{4})}{sin(\frac{\pi}{4})}=\frac{\frac{\sqrt2}{2}}{\frac{\sqrt2}{2}}=1$

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Example Question #26 : Trigonometry

Evaluate: sec(60)-csc(45)

Possible Answers:

$2-\frac{\sqrt2}{2}$

$2\sqrt2+2$

$2+\sqrt2$

$2+\frac{\sqrt2}{2}$

$2-\sqrt2$

Correct answer:

$2-\sqrt2$

Explanation:

Evaluate each term separately.

$sec(60)= \frac{1}{cos(60)}=\frac{1}{0.5}= 2$

$csc(45)= \frac{1}{sin(45)}= \frac{1}{\frac{\sqrt2}{2}}= \frac{2}{\sqrt2}\times\frac{\sqrt2}{\sqrt2}= \frac{2\sqrt2}{2}=\sqrt2$

$sec(60)-csc(45)=2-\frac{2\sqrt2}{2}=2-\sqrt2$

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Example Question #1 : Sec, Csc, Ctan

Pick the ratio of side lengths that would give sec C.

Possible Answers:

$\sec C=\frac{AB}{AC}$

$\sec C=\frac{CB}{AC}$

$\sec C=\frac{AC}{CB}$

None of the other answers.

$\sec C=\frac{AC}{AB}$

Correct answer:

$\sec C=\frac{CB}{AC}$

Explanation:

$\sec\theta=\frac{1}{\cos\theta }$

Find the ratio of Cosine and take the reciprocal.

$\cos\theta=\frac{AC}{CB}\rightarrow \sec\theta=\frac{1}{\frac{AC}{CB}}=\mathbf{\frac{CB}{AC}}$

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Example Question #21 : Trigonometry

Evaluate: sec(30)-csc(60)

Possible Answers:

$\sqrt3-1$

$\frac{\sqrt3-1}{2}$

$\sqrt{3}$

$\frac{1-\sqrt3}{2}$

Correct answer:

Explanation:

Recall that $sec(x) = \frac{1}{cos(x)}$ and $csc(x) = \frac{1}{sin(x)}$ .

Rewrite the expression.

$sec(30)-csc(60) = \frac{1}{cos(30)}-\frac{1}{sin(60)}$

The value of $cos(30) = \frac{\sqrt3}{2}$ and $sin(60) =\frac{ \sqrt{3}}{2}$ .

Since these values are similar, our resulting answer is zero upon substitution.

$\frac{1}{\frac{\sqrt3}{2}}-\frac{1}{\frac{\sqrt3}{2}} = 0$

The answer is:

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Example Question #21 : Trigonometry

Aas

Find the area of the triangle.

Possible Answers:

$6\sqrt{3} + 6$

$18\sqrt{3} + 18$

$12\sqrt{3} + 12$

$12\sqrt{3}$

$6\sqrt{3} + 12$

Correct answer:

$18\sqrt{3} + 18$

Explanation:

Aas_key

Dropping the altitude creates two special right triangles as shown in the diagram. Use the area formula of a triangle to get

$A = \frac {1}{2}bh = \frac {1}{2} \cdot (6 \sqrt{3} + 6) \cdot 6 = 18\sqrt{3} + 18$

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Example Question #21 : Trigonometry

Fire tower A is miles due west of fire tower B. Fire tower A sees a fire in the direction degrees west of north. Fire tower B sees the same fire in the direction degrees east of north. Which tower is closer to the fire and by how much?

Possible Answers:

Fire tower B; 0.24 miles

Fire tower A; 1.53 miles

Fire tower B; 1.29 miles

The two fire towers are equidistant to the fire.

Fire tower A; 0.24 miles

Correct answer:

Fire tower B; 0.24 miles

Explanation:

Fire

First, realize that the angles given are from due north, which means you need to find the complements to find the interior angles of the triangle. This triangle happens to be a right triangle, so the fast way to compute the distances is using trigonometry.

$\sin 50^\circ = \frac{AF}{2}$

$AF \approx 1.53 miles$

$\sin 40^\circ = \frac{BF}{2}$

$BF \approx 1.29 miles$

Fire tower B is 1.53 - 1.29 = 0.24 miles closer to the fire.

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SAT II Math I : Trigonometry

Study concepts, example questions & explanations for SAT II Math I

All SAT II Math I Resources

Example Questions

Example Question #1 : Secant, Cosecant, Cotangent

Example Question #21 : Trigonometry

Example Question #21 : Trigonometry

Example Question #1 : Sec, Csc, Ctan

Example Question #25 : Trigonometry

Example Question #26 : Trigonometry

Example Question #1 : Sec, Csc, Ctan

Example Question #21 : Trigonometry

Example Question #21 : Trigonometry

Example Question #21 : Trigonometry

All SAT II Math I Resources

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