A triangle has sides A, B, and C. The angle between sides A and B is #(pi)/6#. If side C has a length of #1 # and the angle between sides B and C is #(7pi)/12#, what are the lengths of sides A and B?

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Answer 1 · 2017-04-23T04:54:06

#a = (sqrt2+sqrt6)/2, b=sqrt2#

#h_b = c sin(pi - (7pi)/12) = sin((5pi)/12) = (sqrt2+sqrt6)/4#

#a = h_b/sin(pi/6) = ((sqrt2+sqrt6)/4)/(1/2) = (sqrt2+sqrt6)/2#

Angle between #a# and #c# is #pi-pi/6-(7pi)/12 = 3pi/12 = pi/4#

#b^2 = a^2+c^2-2ac cos(pi/4) = (8+4sqrt3)/4 + 1 - (sqrt2+sqrt6)sqrt2/2#

#b^2 = 2+sqrt3 -(2sqrt3)/2 = 2#

#b = sqrt2#

Answer 2 · 2017-04-23T10:09:33

# a=(sqrt(2)+sqrt(6))/2#

#b=sqrt(2)#

We can use the sine rule:

# a/sinA=b/sinB=c/sinC #

So we have:

# A = (7pi)/12; C=pi/6; c==1 #

To find #a# we use

# a/sinA=c/sinC => a/sin((7pi)/12)=1/sin(pi/6) #
# :. a=sin((7pi)/12)/sin(pi/6) = (sqrt(2)+sqrt(6))/2#

To find #b# we use

# A+B+C=pi=>B=pi-(7pi)/12-pi/6=pi/4#

And as before applying thee sin rule gives:

# b/sinB=c/sinC => b/sin(pi/4)=1/sin(pi/6) #
# :. b=sin(pi/4)/sin(pi/6) =sqrt(2)#