A triangle has sides A, B, and C. Sides A and B are of lengths #1# and #2#, respectively, and the angle between A and B is #(pi)/8 #. What is the length of side C?

Question

A triangle has sides A, B, and C.  Sides A and B are of lengths #1# and #2#, respectively, and the angle between A and B is #(pi)/8 #. What is the length of side C?

Quinn Anderson · Accepted Answer

Answer for the updated question given below.
#c~~1.14# #color(red)( triangle ABC)# has sides#color(red)(AB, BC,and CA#. Sides #color(red)(a)# and #color(red)(b)# are of lengths 1 and 2 respectively, and the angle between #color(red)(a)# and #color(red)(b)# is #pi/8#. What is the length of side #color(red)(c)#? Where#color(red)(AB=c,BC=a and CA=b.)# Here, #a=1,b=2,C=pi/8# Using the Law of Cosines: #cosC=(a^2+b^2-c^2)/(2ab) rArrc^2=a^2+b^2-2abcosC# Hence,#c^2=(1)^2+(2)^2-2(1)(2)cos(pi/8) ~~1+4-4(0.9239) rArr c^2~~5-3.6956=1.3044##color(red)(rArrc~~1.14)#

Josiah Anderson · Answer

undefined Using the Law of Cosines:

$C^2 = A^2 + B^2 - 2AB \cdot \cos(C)$

Substitute the given values:

$C^2 = 1^2 + 2^2 - 2 \cdot 1 \cdot 2 \cdot \cos(\pi/8)$

Calculate the cosine of $ \pi/8 $:

$ \cos(\pi/8) \approx 0.9239 $

Substitute into the equation:

$C^2 = 1 + 4 - 4 \cdot 0.9239$

$C^2 ≈ 1 + 4 - 3.6956$

$C^2 ≈ 1.3044$

$C ≈ \sqrt{1.3044}$

$C ≈ 1.1412$

So, the length of side C is approximately $1.1412$.