A triangle has sides A, B, and C. The angle between sides A and B is #(7pi)/12#. If side C has a length of #1 # and the angle between sides B and C is #pi/12#, what is the length of side A?

1 Answer
Lotusbluete
Jan 25, 2016

The length of side #A# is #2 - sqrt(3) #.

Explanation:

You can use the law of sines.

The angle between sides #A# and #B# is the angle opposing the side #C#. Let's call this angle #gamma#.

Similarly, the angle between sides #B# and #C# is the one opposing the side #A#. Let's call this angle #alpha#.

Even though we don't need it for this particular task, let's also call the last remaining angle #beta#, this one is opposing the side #B# .

According to the law of sines, the following relation between the sides and the opposite angles exists:

#A / sin alpha = B / sin beta = C / sin gamma#

In our case, we only need to look at #A#, #C#, #alpha# and #gamma#:

#A / sin alpha = C / sin gamma#

#A / sin (pi / 12) = 1 / sin ((7 pi)/12)#

#<=> A = sin (pi / 12) / sin ((7pi) / 12)#

So, the only thing left to do is compute #sin (pi/12)# and #sin((7pi)/12)#.

Let me show you how to do this without the calculator but with some basic knowledge about #sin# and #cos# and using the identity

#sin (a-b) = sin(a)cos(b) - cos(a)sin(b)#.

You need to express #pi/12# as a sum or difference of simpler values:

#sin (pi/12) = sin (pi/3 - pi/4)#

# = sin(pi/3) cos(pi/4) - cos(pi/3)sin(pi/4)#

# = sqrt(3)/2 * 1/sqrt(2) - 1 / 2 * 1 / sqrt(2)#

# = 1/(2 sqrt(2)) (sqrt(3) - 1)#

Similarly,

#sin((7 pi)/12) = sin(pi/3 + pi/4)#

# = sin(pi/3) cos(pi/4) + cos(pi/3) sin(pi/4)#

# = sqrt(3) / 2 * sqrt(2) / 2 + 1 / 2 * sqrt(2) / 2#

# = 1/(2 sqrt(2)) (sqrt(3) + 1)#

Thus, the length of the side #A# is

# A = sin (pi / 12) / sin ((7pi) / 12)#

# = (1/(2 sqrt(2)) (sqrt(3) - 1) )/(1/(2 sqrt(2)) (sqrt(3) + 1))#

# = (sqrt(3) - 1) / (sqrt(3) + 1)#

# = ((sqrt(3) - 1) color(blue)((sqrt(3)-1))) / ((sqrt(3) + 1) color(blue)((sqrt(3)-1)))#

# = (sqrt(3) -1)^2 / ((sqrt(3))^2 - 1^2 ) #

# = (3 - 2sqrt(3) + 1) / 2 #

# = 2 - sqrt(3) #

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