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?

Apr 23, 2017

$a = \frac{\sqrt{2} + \sqrt{6}}{2} , b = \sqrt{2}$

Explanation:

${h}_{b} = c \sin \left(\pi - \frac{7 \pi}{12}\right) = \sin \left(\frac{5 \pi}{12}\right) = \frac{\sqrt{2} + \sqrt{6}}{4}$

$a = {h}_{b} / \sin \left(\frac{\pi}{6}\right) = \frac{\frac{\sqrt{2} + \sqrt{6}}{4}}{\frac{1}{2}} = \frac{\sqrt{2} + \sqrt{6}}{2}$

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

${b}^{2} = {a}^{2} + {c}^{2} - 2 a c \cos \left(\frac{\pi}{4}\right) = \frac{8 + 4 \sqrt{3}}{4} + 1 - \left(\sqrt{2} + \sqrt{6}\right) \frac{\sqrt{2}}{2}$

${b}^{2} = 2 + \sqrt{3} - \frac{2 \sqrt{3}}{2} = 2$

$b = \sqrt{2}$

Apr 23, 2017

$a = \frac{\sqrt{2} + \sqrt{6}}{2}$

$b = \sqrt{2}$

Explanation:

We can use the sine rule:

$\frac{a}{\sin} A = \frac{b}{\sin} B = \frac{c}{\sin} C$

So we have:

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

To find $a$ we use

$\frac{a}{\sin} A = \frac{c}{\sin} C \implies \frac{a}{\sin} \left(\frac{7 \pi}{12}\right) = \frac{1}{\sin} \left(\frac{\pi}{6}\right)$
$\therefore a = \sin \frac{\frac{7 \pi}{12}}{\sin} \left(\frac{\pi}{6}\right) = \frac{\sqrt{2} + \sqrt{6}}{2}$

To find $b$ we use

$A + B + C = \pi \implies B = \pi - \frac{7 \pi}{12} - \frac{\pi}{6} = \frac{\pi}{4}$

And as before applying thee sin rule gives:

$\frac{b}{\sin} B = \frac{c}{\sin} C \implies \frac{b}{\sin} \left(\frac{\pi}{4}\right) = \frac{1}{\sin} \left(\frac{\pi}{6}\right)$
$\therefore b = \sin \frac{\frac{\pi}{4}}{\sin} \left(\frac{\pi}{6}\right) = \sqrt{2}$