1. ## Law of Sines

Two triangles can be formed with the given information. Use the Law of Sines to solve the triangles.

B = 49°, a = 16, b = 14

Choices are

A = 30.4°, C = 100.6°, c = 18.2; A = 149.6°, C = 79.4°, c = 18.2
A = 30.4°, C = 100.6°, c = 10.7; A = 149.6°, C = 79.4°, c = 10.7
A = 59.6°, C = 71.4°, c = 11.1; A = 120.4°, C = 10.6°, c = 11.1
A = 59.6°, C = 71.4°, c = 17.6; A = 120.4°, C = 10.6°, c = 3.4

2. ## Re: Law of Sines

Do you know what the Law of Sines is?

With the information you have been given, you should be able to calculate A.
Then you should be able to calculate C, and c.

Not sure how you can get another triangle from that information.

3. ## Re: Law of Sines

Hello, ambitionty9!

Two triangles can be formed with the given information.
Use the Law of Sines to solve the triangles.

. . $B = 49^o,\;a=16,\;b=14$

$[1]\;A = 30.4^o,\;C = 100.6^o,\;c = 18.2$
. . $A = 149.6^o,\;C = 79.4^o,\;c = 18.2$

$[2]\;A = 30.4^o,\;C = 100.6^o,\;c = 10.7$
. . $A = 149.6^o,\;C = 79.4^o,\;c = 10.7$

$[3]\;A = 59.6^o,\;C = 71.4^o,\;c = 11.1$
. . $A = 120.4^o,\;C = 10.6^o,\;c = 11.1$

$[4]\;A = 59.6^o,\;C = 71.4^o,\;c = 17.6$
. . $A = 120.4^o,\;C = 10.6^o,\;c = 3.4$

Law of Sines: . $\frac{\sin A}{a} \,=\,\frac{\sin B}{b} \quad\Rightarrow\quad \sin A \,=\,\frac{a\sin B}{b}$

We have: . $\sin A \:=\:\frac{16\sin49^o}{14} \:=\:0.862525235$

Hence: . $A \:=\:59.60130951 \quad\Rightarrow\quad \boxed{A \:\approx\:59.6^o}$ . **

Then: . $C \:=\:180^o - 49^o - 59.6^o \quad\Rightarrow\quad \boxed{C \:=\:71.4^o}$

Law of Sines: . $\frac{c}{\sin C} \,=\,\frac{b}{\sin B} \quad\Rightarrow\quad c \:=\:\frac{b\sin C}{\sin B}$

We have: . $c \:=\:\frac{14\sin71.4^o}{\sin49^o} \:=\:17.58127641 \quad\Rightarrow\quad \boxed{c \:\approx\:17.6}$

** .We also have: . $\boxed{A \:=\:120.4^o}$

Then: . $C \:=\:180^o - 49^o - 120.4^o \quad\Rightarrow\quad \boxed{C \:=\:10.6^o}$

We have: . $c \:=\:\frac{14\sin10.6^o}{\sin29^o} \:=\:3.412331016 \quad\Rightarrow\quad \boxed{c \:=\:3.4}$