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SAT II Math II : SAT Subject Test in Math II

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

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All SAT II Math II Resources

2 Diagnostic Tests 130 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #461 : Sat Subject Test In Math Ii

The degree angle 300 can be expressed as what in radians?

Possible Answers:

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

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

$\frac{4}{3}\pi$

$\frac{5}{3}\pi$

$\frac{6}{5}\pi$

Correct answer:

$\frac{5}{3}\pi$

Explanation:

In order to convert degrees to radians, we will need to know the conversion factor.

$180 \textup{ deg} = \pi \textup{ rad}$

Set up a dimensional analysis to solve.

$300 \textup{ deg} (\frac{\pi \textup{ rad}}{180 \textup{ deg}}) = \frac{5}{3}\pi\textup{ rad}$

The answer is: $\frac{5}{3}\pi$

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Example Question #1 : Radians And The Unit Circle

Which of the following angles belong in the fourth quadrant?

Possible Answers:

$\frac{4}{3}\pi$

$\frac{17}{9}\pi$

$\pi$

$\frac{5}{2}\pi$

Correct answer:

$\frac{17}{9}\pi$

Explanation:

The fourth quadrant is in the positive x-axis and negative y-axis.

The angle ranges are:

$270\rightleftharpoons 360 \textup{ deg}$

$\frac{3}{2}\pi\rightleftharpoons 2\pi \textup{ radians}$

The only possible answer is: $\frac{17}{9}\pi$

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Example Question #461 : Sat Subject Test In Math Ii

What degree measure is equivalent to $4\pi$ ?

Possible Answers:

540

720

1020

980

360

Correct answer:

720

Explanation:

Every pi radians is equal to 180 degrees.

Replace the pi term with 180 degrees and multiply.

$4\pi = 4(180 \textup{ degrees}) = 720$

The answer is: 720

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Example Question #1 : Law Of Cosines

A triangle has sides that measure 10, 12, and 16. What is the greatest measure of any of its angles (nearest tenth of a degree)?

Possible Answers:

$87.1^{\circ}$

$141.4^{\circ}$

$92.9^{\circ}$

$131.5^{\circ}$

$120.0^{\circ}$

Correct answer:

$92.9^{\circ}$

Explanation:

We are seeking the measure of the angle opposite the side of greatest length, 16.

We can use the Law of Cosines, setting c = 16, a= 10, b = 12 , and solving for :

$\cos C = \frac{a^{2}+b^{2}- c^{2}}{2ab}$

$\cos C = \frac{10^{2}+12^{2}- 16^{2}}{2 \cdot 10 \cdot 12}$

$\cos C = \frac{100+144- 256}{240}$

$\cos C = \frac{-12}{240}$

$\cos C =- 0.0500$

$C = \cos ^{-1}(-0.0500) \approx 92.9^{\circ}$

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Example Question #462 : Sat Subject Test In Math Ii

A triangle has sides that measure 15, 17, and 30. What is the least measure of any of its angles (nearest tenth of a degree)?

Possible Answers:

$70.9^{\circ}$

$67.8^{\circ}$

$68.3^{\circ}$

$21.7^{\circ}$

$19.1^{\circ}$

Correct answer:

$19.1^{\circ}$

Explanation:

We are seeking the measure of the angle opposite the side of least length, 15.

We can use the Law of Cosines, setting c = 15, a= 17, b = 30 , and solving for :

$\cos C = \frac{a^{2}+b^{2}- c^{2}}{2ab}$

$\cos C = \frac{17^{2}+30^{2}- 15^{2}}{2 \cdot 17 \cdot 30}$

$\cos C = \frac{289+900- 225 }{1,020}$

$\cos C = \frac{964 }{1,020}$

$\cos C \approx 0.9451$

$C \approx \cos ^{-1} 0.9451 \approx 19.1^{\circ}$

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Example Question #463 : Sat Subject Test In Math Ii

Given : $\Delta ABC$ with $AB = 12, BC = 17, m \angle B= 136 ^{\circ}$ .

Which of the following whole numbers is closest to ?

Possible Answers:

Correct answer:

Explanation:

Apply the Law of Cosines

$c =\sqrt{ a^{2}+b^{2}- 2ab \cos C}$

setting $a = 12, b= 17, C= 136 ^{\circ}$ and solving for :

$c =\sqrt{ 12^{2}+17^{2}- 2\cdot 12 \cdot 17 \cos 136^{\circ}}$

$c \approx \sqrt{144+289-408 (-0.7193)}$

$c \approx \sqrt{144+289+293.49}$

$c \approx \sqrt{726.49}$

$c \approx 27.0$

Of the five choices, 27 comes closest.

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Example Question #4 : Law Of Cosines

Given : $\Delta ABC$ with $AB = 12, BC = 16, m \angle B= 56 ^{\circ}$ .

Evaluate to the nearest tenth.

Possible Answers:

14.7

13.6

9.0

13.0

The correct answer is not given among the other responses.

Correct answer:

13.6

Explanation:

Apply the Law of Cosines

$c =\sqrt{ a^{2}+b^{2}- 2ab \cos C}$

setting $a = 12, b = 16, C = 56^{\circ}$ and solving for :

$c =\sqrt{ 12^{2}+16^{2}- 2\cdot 12 \cdot 16 \cos 56^{\circ}}$

$c \approx \sqrt{ 144+256- 384 \cdot 0.5592 }$

$c \approx \sqrt{ 144+256- 214.73 }$

$c \approx \sqrt{ 185.27 }$

$c \approx 13.6$

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Example Question #1 : Law Of Cosines

In $\bigtriangleup ABC$ :

AB = 12

AC = 15

$m \angle A = 105^{\circ }$

Evaluate the length of $\overline{BC}$ to the nearest tenth of a unit.

Possible Answers:

21.2

26.8

16.6

21.6

19.2

Correct answer:

21.6

Explanation:

The figure referenced is below:

Triangle 2

By the Law of Cosines, given the lengths and of two sides of a triangle, and the measure $\gamma$ of their included angle, the length of the third side can be calculated using the formula

$c^{2} = a ^{2} + b^{2} - 2ab \cos \gamma$

Substituting a = AB = 12 , b = AC = 15 , c = BC , and $\gamma = m \angle A = 105^{\circ }$ , then evaluating:

$(BC)^{2} = 12^{2} + 15 ^{2} - 2 (12)(15)\cos 105^{\circ }$

$\approx 144+225 - 360 (-0.2588 )$

$\approx 369 - (-93.17 )$

$\approx 462.17$

Taking the square root of both sides:

$BC\approx \sqrt{462.17} \approx 21.6$

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Example Question #1 : Law Of Sines

Find the measure of angle .

Possible Answers:

61.20

70.27

78.61

109.73

Correct answer:

70.27

Explanation:

Start by using the Law of Sines to find the measure of angle .

$\frac{8}{\sin 39}=\frac{12}{\sin B}$

$12\sin 39=8\sin B$

$\sin B=0.94398$

$B=70.73^{\circ}$

Since the angles of a triangle must add up to 180 ,

39+70.73+A=180

$A=70.27^{\circ}$

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Example Question #1 : Law Of Sines

In $\bigtriangleup ABC$ ,

AB = 23

BC = 34

$\angle C =65 ^{\circ }$

Evaluate $m \angle A$ (nearest degree)

Possible Answers:

$38^{\circ }$

$41^{\circ }$

$74^{\circ }$

Cannot be determined

$77^{\circ }$

Correct answer:

Cannot be determined

Explanation:

By the Law of Sines, if and are the lengths of two sides of a triangle, and $\alpha$ and $\beta$ the measures of their respective opposite angles,

$\frac{\sin \alpha }{a} = \frac{\sin \beta }{b}$

$\angle A$ and $\angle C$ are opposite sides $\overline{BC}$ and $\overline{AB}$ , so, setting $\alpha = m \angle A$ , $\beta = m\angle C = 65^{\circ }$ , a = BC = 34 , and b = AB = 23 :

$\frac{\sin \alpha }{34} = \frac{\sin 65^{\circ } }{23}$

$\frac{\sin \alpha }{34}\cdot 34 = \frac{\sin 65^{\circ } }{23} \cdot 34$

$\sin \alpha \approx \frac{0.9063 }{23} \cdot 34 \approx 1.3398$

However, the range of the sine function is [-1, 1] , so there is no value of $\alpha$ for which this is true. Therefore, this triangle cannot exist.

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SAT II Math II : SAT Subject Test in Math II

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

All SAT II Math II Resources

Example Questions

Example Question #461 : Sat Subject Test In Math Ii

Example Question #1 : Radians And The Unit Circle

Example Question #461 : Sat Subject Test In Math Ii

Example Question #1 : Law Of Cosines

Example Question #462 : Sat Subject Test In Math Ii

Example Question #463 : Sat Subject Test In Math Ii

Example Question #4 : Law Of Cosines

Example Question #1 : Law Of Cosines

Example Question #1 : Law Of Sines

Example Question #1 : Law Of Sines

All SAT II Math II Resources

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