A triangle has sides A, B, and C. Sides A and B are of lengths 2 and 5, respectively, and the angle between A and B is pi/12. What is the length of side C?

Jan 27, 2016

$C \approx 4.4$

Explanation: Use the trigonometric relationships e.g. sine = opposite /hypotenuse
$A = x + y$

$\frac{x}{B} = \cos \left(\frac{\pi}{12}\right)$

$\therefore x = B \cos \left(\frac{\pi}{12}\right) = 5 \cos \left(\frac{\pi}{12}\right)$

Similarly $h = 5 \sin \left(\frac{\pi}{12}\right)$

$y = A - x = 2 - 5 \cos \left(\frac{\pi}{12}\right)$

Using Pythagoras, ${C}^{2} = {h}^{2} + {y}^{2}$

:.${C}^{2} = {\left(5 \sin \left(\frac{\pi}{12}\right)\right)}^{2} + {\left(2 - 5 \cos \left(\frac{\pi}{12}\right)\right)}^{2}$

${C}^{2} = 25 {\sin}^{2} \left(\frac{\pi}{12}\right) + 4 - 10 \cos \left(\frac{\pi}{12}\right) + 25 {\cos}^{2} \left(\frac{\pi}{12}\right)$

${C}^{2} = 25 \left({\sin}^{2} \left(\frac{\pi}{12}\right) + {\cos}^{2} \left(\frac{\pi}{12}\right)\right) + 4 - 10 \cos \left(\frac{\pi}{12}\right)$

Using the fact that ${\sin}^{2} \left(\theta\right) + {\cos}^{2} \left(\theta\right) = 1$ this is now

${C}^{2} = 25 + 4 - 10 \cos \left(\frac{\pi}{12}\right) = 29 - 10 \cos \left(\frac{\pi}{12}\right)$

$C \approx \sqrt{29 - 9.659} \approx 4.4$