# Trigonometry/Solving Triangles Given SAS

## Given Two Sides and the Included Angle (SAS)Edit

Given two sides and the angle included by the two given sides, we can apply the Law of Cosines to find the missing side.

 Spelling out the detail Let's say the sides we know are $\displaystyle b$ and $\displaystyle c$ and the included angle is $\displaystyle \alpha$ We want to find the missing side $\displaystyle a$ We know: $b^2+c^2-2bc\cos{\alpha}=a^2, \,$ We have $\displaystyle b, c$ and $\displaystyle \alpha$. We can therefore work out the missing side $\displaystyle a$ as $a = \sqrt{b^2+c^2-2bc\cos{\alpha}} \,$ Don't be confused by the different letters we are using here to the original statement of the law of cosines! Think of the law of cosines as being a relation between two sides and the included angle, to the remaining side. We have many choices as to how we label the sides and angles.

At this point we will know a side and an opposite angle. That's where the law of sines comes in. It says that in any triangle the ratio of the length of a side to the sine of the opposite angle is the same for all three sides. We can therefore use the law of sines to find the sine of the angle opposite each side that we know. Then, knowing the sine of the angle we can work out the angle itself.

 Spelling out the detail We know the ratio: $\displaystyle \frac{\sin\alpha}{a}$ is the same for the other angles and sides. So in particular: $\displaystyle \sin \gamma = \sin\alpha \frac{c}{a}$ and $\displaystyle \sin \beta = \sin\alpha \frac{b}{a}$ We know the quantities on the right. We can calculate the quantities on the left from these. Take the inverse sines and we're done. Not quite. We need to check that the angles $\displaystyle \alpha, \beta$ and $\displaystyle \gamma$ add up to 180o. You might be horrified sometime to find that they don't. What can go wrong? See the next box.
 Ambiguity of inverse sin What is $\displaystyle \sin 80^\circ$? What is $\displaystyle \sin 100^\circ$? (you are expected to use a calculator for these). Do you see the problem? The inverse sine function 'makes the wrong choice' when doing the inverse sine of an obtuse angle. So, in solving an ASA triangle, you need to spot if one of the angles is obtuse. If the inverse sine for it tells you an acute angle you can then correct it to the equivalent obtuse angle. $\displaystyle \sin 70^\circ = \sin 110^\circ$ $\displaystyle \sin 60^\circ = \sin 120^\circ$ $\displaystyle \sin 20^\circ = \sin 160^\circ$ The angles with the same sine add up to 180, so you can correct the angle that came out acute that should be obtuse by subtracting it from 180o. Alternatively, since at most one of the angles in a triangle can be obtuse, the one opposite the longest side, calculate the other angle instead. Then use the fact that the angles sum to 180o to get the remaining one.

As before, the area can be found from Heron's Formula.