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

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?

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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.

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Thanks! :D ah 180 ! – user1022734 Feb 6 '12 at 12:52
There's a second solution that you didn't account for. – Isaac Feb 6 '12 at 16:37

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.

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Can't believe I missed that. – Lazar Ljubenović Feb 6 '12 at 16:54

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 :

$$\gamma = 180° - (\alpha + \beta)$$

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There's a second solution that you didn't account for. – Isaac Feb 6 '12 at 16:37

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.

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There's a second solution that you didn't account for. – Isaac Feb 6 '12 at 16:37