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?

2 Answers
Apr 23, 2017

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

Explanation:

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

Apr 23, 2017

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

b=sqrt(2)

Explanation:

We can use the sine rule:

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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)