# Law of Sines

• May 14th 2013, 05:05 PM
ambitionty9
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
• May 16th 2013, 11:22 AM
DavidB
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.
• May 16th 2013, 08:59 PM
Soroban
Re: Law of Sines
Hello, ambitionty9!

Quote:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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