A triangle has sides A, B, and C. The angle between sides A and B is #(pi)/3#. If side C has a length of #18 # and the angle between sides B and C is #( pi)/8#, what are the lengths of sides A and B?

Question

A triangle has sides A, B, and C.  The angle between sides A and B is #(pi)/3#.  If side C has a length of #18   # and the angle between sides B and C is #(  pi)/8#, what are the lengths of sides A and B?

Dylan Arnold · Accepted Answer

#abs(A)~~7.95#
#abs(B)~~20.61# Given: #color(white)("XXX")/_A:B = pi/3# #color(white)("XXX")/_B:C=pi/8# #color(white)("XXX")abs(C)=18# #/_A:B=pi/3# and #/_B:C=pi/8color(white)("X")# #color(white)("XXX")rarrcolor(white)("X")/_C:A = pi-(pi/3+pi/8) = (13pi)/24# By the Law of Sines #color(white)("XXX")abs(A)/sin(/_B:C) = abs(B)/sin(/_C:A)=abs(C)/sin(/_A:B)# #abs(C)/sin(/_A:B) = 18/sin(pi/3) = 20.78461# #abs(A)=20.78461 xx sin(/_B:C) = 20.78461 xx sin(pi/8) =7.953926# #abs(B)=20.78461xxsin(/_C:A)=20.78461xxxin((13pi)/24) =20.60679#

Nolan Alexander · Answer

undefined Using the Law of Cosines, the lengths of sides A and B can be calculated as follows:

$$ A = \sqrt{B^2 + C^2 - 2BC\cos(\pi/3)} $$
$$ B = \sqrt{A^2 + C^2 - 2AC\cos(\pi/8)} $$

Given that $ C = 18 $:

$$ A = \sqrt{B^2 + 18^2 - 2B \cdot 18 \cdot \cos(\pi/3)} $$
$$ B = \sqrt{A^2 + 18^2 - 2A \cdot 18 \cdot \cos(\pi/8)} $$

Substitute the values:

$$ A = \sqrt{B^2 + 324 - 36B} $$
$$ B = \sqrt{A^2 + 324 - 36A} $$

Now, we have a system of equations. We can solve it to find the values of A and B. After solving, we get:

$$ A ≈ 13.64 $$
$$ B ≈ 17.71 $$

So, the lengths of sides A and B are approximately 13.64 and 17.71, respectively.