Law of Sines - SAT Math

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Question

In $\bigtriangleup ABC$ ,

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

Evaluate $m \angle A$ (nearest degree)

Answer

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 $\alpha = m \angle A$ , $\beta = m\angle C = 65^{\circ }$ , , and :

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

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

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

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

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Question

Find the measure of angle .

Answer

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 ,

$A=70.27^{\circ}$

Compare your answer with the correct one above

Question

In $\bigtriangleup ABC$ ,

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

Evaluate $m \angle A$ (nearest degree)

Answer

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 $\alpha = m \angle A$ , $\beta = m\angle C = 65^{\circ }$ , , and :

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

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

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

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

Compare your answer with the correct one above

Question

Find the measure of angle .

Answer

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 ,

$A=70.27^{\circ}$

Compare your answer with the correct one above

Question

In $\bigtriangleup ABC$ ,

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

Evaluate $m \angle A$ (nearest degree)

Answer

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 $\alpha = m \angle A$ , $\beta = m\angle C = 65^{\circ }$ , , and :

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

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

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

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

Compare your answer with the correct one above

Question

Find the measure of angle .

Answer

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 ,

$A=70.27^{\circ}$

Compare your answer with the correct one above

Question

In $\bigtriangleup ABC$ ,

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

Evaluate $m \angle A$ (nearest degree)

Answer

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 $\alpha = m \angle A$ , $\beta = m\angle C = 65^{\circ }$ , , and :

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

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

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

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

Compare your answer with the correct one above

Question

Find the measure of angle .

Answer

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 ,

$A=70.27^{\circ}$

Compare your answer with the correct one above

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