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SAT II Math II : Law of Sines

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

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Example Questions

Example Question #461 : Sat Subject Test In Math Ii

Find the measure of angle $\displaystyle A$.

Possible Answers:

109.73 $\displaystyle 109.73$

70.27 $\displaystyle 70.27$

61.20 $\displaystyle 61.20$

78.61 $\displaystyle 78.61$

Correct answer:

70.27 $\displaystyle 70.27$

Explanation:

Start by using the Law of Sines to find the measure of angle $\displaystyle B$.

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

$12\sin 39=8\sin B$ $\displaystyle 12\sin 39=8\sin B$

$\sin B=0.94398$ $\displaystyle \sin B=0.94398$

$B=70.73^{\circ}$ $\displaystyle B=70.73^{\circ}$

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

$\displaystyle 39+70.73+A=180$

$A=70.27^{\circ}$ $\displaystyle A=70.27^{\circ}$

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Example Question #462 : Sat Subject Test In Math Ii

In $\bigtriangleup ABC$ $\displaystyle \bigtriangleup ABC$,

AB = 23 $\displaystyle AB = 23$

BC = 34 $\displaystyle BC = 34$

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

Evaluate $m \angle A$ $\displaystyle m \angle A$ (nearest degree)

Possible Answers:

Cannot be determined

$77^{\circ }$ $\displaystyle 77^{\circ }$

$41^{\circ }$ $\displaystyle 41^{\circ }$

$38^{\circ }$ $\displaystyle 38^{\circ }$

$74^{\circ }$ $\displaystyle 74^{\circ }$

Correct answer:

Cannot be determined

Explanation:

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

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

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

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

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

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

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

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SAT II Math II : Law of Sines

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 #462 : Sat Subject Test In Math Ii

All SAT II Math II Resources

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