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

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

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Answer 1 · 2017-04-23T04:54:06

$a = \frac{\sqrt{2} + \sqrt{6}}{2}, b = \sqrt{2}$ $a = (sqrt2+sqrt6)/2, b=sqrt2$

Explanation:

$h_{b} = c sin (π - \frac{7 π}{12}) = sin (\frac{5 π}{12}) = \frac{\sqrt{2} + \sqrt{6}}{4}$ $h_b = c sin(pi - (7pi)/12) = sin((5pi)/12) = (sqrt2+sqrt6)/4$

$a = \frac{h_{b}}{sin (\frac{π}{6})} = \frac{\frac{\sqrt{2} + \sqrt{6}}{4}}{\frac{1}{2}} = \frac{\sqrt{2} + \sqrt{6}}{2}$ $a = h_b/sin(pi/6) = ((sqrt2+sqrt6)/4)/(1/2) = (sqrt2+sqrt6)/2$

Angle between $a$ $a$ and $c$ $c$ is $π - \frac{π}{6} - \frac{7 π}{12} = 3 \frac{π}{12} = \frac{π}{4}$ $pi-pi/6-(7pi)/12 = 3pi/12 = pi/4$

$b^{2} = a^{2} + c^{2} - 2 a c cos (\frac{π}{4}) = \frac{8 + 4 \sqrt{3}}{4} + 1 - (\sqrt{2} + \sqrt{6}) \frac{\sqrt{2}}{2}$ $b^2 = a^2+c^2-2ac cos(pi/4) = (8+4sqrt3)/4 + 1 - (sqrt2+sqrt6)sqrt2/2$

$b^{2} = 2 + \sqrt{3} - \frac{2 \sqrt{3}}{2} = 2$ $b^2 = 2+sqrt3 -(2sqrt3)/2 = 2$

$b = \sqrt{2}$ $b = sqrt2$

Answer 2 · 2017-04-23T10:09:33

$a = \frac{\sqrt{2} + \sqrt{6}}{2}$ $a=(sqrt(2)+sqrt(6))/2$

$b = \sqrt{2}$ $b=sqrt(2)$

Explanation:

We can use the sine rule:

$\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$ $a/sinA=b/sinB=c/sinC$

So we have:

$A = \frac{7 π}{12}; C = \frac{π}{6}; c = = 1$ $A = (7pi)/12; C=pi/6; c==1$

To find $a$ $a$ we use

$\frac{a}{sin A} = \frac{c}{sin C} \Rightarrow \frac{a}{sin (\frac{7 π}{12})} = \frac{1}{sin (\frac{π}{6})}$ $a/sinA=c/sinC => a/sin((7pi)/12)=1/sin(pi/6)$
$:. a=sin((7pi)/12)/sin(pi/6) = (sqrt(2)+sqrt(6))/2$

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:

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

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

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