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SAT II Math II : Trigonometry

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

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

Example Questions

Example Question #462 : Sat Subject Test In Math Ii

Find the measure of angle .

Possible Answers:

78.61

61.20

109.73

70.27

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 : Trigonometry

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

All SAT II Math II Resources

Example Questions

Example Question #462 : Sat Subject Test In Math Ii

Example Question #1 : Law Of Sines

All SAT II Math II Resources

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