A triangle has sides A, B, and C. The angle between sides A and B is #(3pi)/4#. If side C has a length of #16 # and the angle between sides B and C is #pi/12#, what are the lengths of sides A and B?

1 Answer
Feb 12, 2016

Side #b = 11.31# and Side #a = 5.86#
:)

Explanation:

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We can get the value of side A and B by using the "Law of Sines",

#sin A/a = Sin B/b = Sin C/c#

First, we must convert the radian value to degree value.

To convert radian value to degree value,

Multiply it by #180/pi#

since Angle #A = pi/12#

#A =pi/12 * 180/pi# = #(180pi)/(12pi)# = #(180cancelpi)/(12cancelpi)#

Angle #A = 15^o#

and Angle #C = (3pi)/4 or 135^o#

#C = (3pi)/4 * 180/pi# = #(540pi)/(4pi)# = #(540cancelpi)/(4cancelpi)#

Angle #C = 135^o#

using the law of sines,

#sin A/a = sin C/c#

#(sin15^o)/a = (sin135^o)/16#

using algebraic technique we get,

#a = ((sin15^o)(16))/(sin135^o)#

#a = 5.8564#

we use again the "Law of Sines", since "Pythagorean Theorem" doesn't work on Non-Right Triangles,

since angle #B# is unknown, we can get its value by simply getting the difference of #180^o-(#Angle#C + #Angle# A)#, since "the sum of all interior angles of a triangle is always #180^o#"

Angle #B = 180^o-##(135^o##+15^o)#

#= 180^o-150^o#

Angle #B = 30^o#

Applying again the Law of Sines to get the value of side of #b#, we get,

#sin B/b = sin C/c#

#(sin30^o)/b = (sin135^o)/16#

Applying algebraic technique, we get,

Side #b = ((sin30^o)(16))/(sin135^o)#

Hence, we get:

Side #b = 11.3137#

Tip on Trigonometry:
Pythagorean Theorem is only reliable in solving right triangles, while Law of Sines and Cosines works in almost any triangles

:)