Solving a triangle given two side lengths and the measure of a non-included angle

Question

Let's say given an angle A = 46 °, side a = 2.29 and b = 2.71

I figured that the angle B = 58.4 by saying:

$$B = \sin^{-1} \left(\frac{ 2.71 \sin{46^{\circ}}}{2.29}\right)=58.4^{\circ}$$

But I think that angle C is incorrect:

$$C = \sin^{-1} \left(\frac{2.29 \sin{58.4^{\circ}}}{2.71}\right)=46.03^{\circ}$$

Someone who can help me? what do I do wrong and how should it be done?

Answer 1

You can find sine of angle $B$ using sine rule: $sin\beta=(2.71\cdot sin46)/2.29=0.85 \Rightarrow \beta=58.35$. You got this correctly. Now, using the fact that sum of angles in triangle is 180 degrees, you get angle $C=180-46-58.35=75.65$ (all angles are in degrees).

Once you have two angles, there's no need to re-use sine rule. Calculation is more complicated and you can make a mistake more easily.

If you want to calculate sides of triangle, use cosine rule.

Answer 2

According to Sine Law :

$$\frac{b}{\sin \beta} =\frac{a}{\sin \alpha} \Rightarrow \beta = \arcsin \left(\frac{b\cdot \sin \alpha}{a}\right)$$

Once you find angle $\beta$ you can calculate $\gamma$ from :

$$\gamma = 180° - (\alpha + \beta)$$

Answer 3

Your application of the sine rule to get $C$ is incorrect. It should be

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

but you wrote

$$\frac{\sin C}{\sin A} \overset{?}{=} \frac{a}{b}$$

The correct angle can be worked out from the sum of angles, as pointed out in the other answers.

Answer 4

While the other answers have covered the basic use of the Law of Sines, they've all missed a critical point: there are two triangles that fit your given information. Here is the triangle you've found:

But, when you're solve $\sin B=\frac{b\sin A}{a}$, there are two solutions that could be angles in a triangle, $B_1=\arcsin\left(\frac{b\sin A}{a}\right)$ (the one you found) and $B_2=180°-\arcsin\left(\frac{b\sin A}{a}\right)$. Using this second possible measure for $B$ won't always yield a triangle, but in this case it does:

Here are both triangles pictured together:

This issue with the Law of Sines, sometimes called the "ambiguous case," is almost certainly what the picture in your question here is about.

See also my answer here on general techniques for triangle-solving.