Solving a triangle, given two sides and the measure of the included angle

Let say you have a triangle

Angle A = 41 degrees , side b = 3.41 and c = 5.83

can you use pythagoras theorem to find the side a? and how can you find Angle B and C

-

There isn't a simple way to use the Pythagorean Theorem to find any of the unknown information about the triangle. It is possible to use some right triangle trigonometry in combination with the Pythagorean Theorem to find the unknown information, but it's simpler to use the Law of Cosines—to find the unknown side and then to find the unknown angles.

-
Any small example? –  user1022734 Feb 5 '12 at 22:01
@user1022734: Use $a^2=b^2+c^2-2bc\cos A$ to find $a$, then use $\cos C=\frac{a^2+b^2-c^2}{2ab}$ to find $C$ and a similar equation to find $B$. –  Isaac Feb 5 '12 at 22:03
to find a i most do : b^2+c^ - 2 *b *cos(A) ? or –  user1022734 Feb 5 '12 at 22:06
square root((((3,41^2) + (5,83^2)) - (2 * 3,41)) * 5,83) * cos(41 degree) = 11,3504535 ? –  user1022734 Feb 5 '12 at 23:29
@user1022734: I think your parentheses are misplaced. I've taken your expression and moved the parentheses, underlined things that are in parentheses together: $$\text{sqrt}\underline{(\underline{(\underline{(3,41^2)} + \underline{(5,83^2)})} - \underline{(\underline{(\underline{(2 * 3,41)} * 5,83)} * \cos\underline{(41^\circ)})})}\approx 3.95086$$ –  Isaac Feb 5 '12 at 23:38

To find side $a$ use Law Of Cosines:

$$a^2=b^2+c^2-2\cdot b \cdot c \cos \alpha$$

To find angles $\beta$ and $\gamma$ use Sine Law :

$$\frac{a}{\sin \alpha} =\frac{b}{\sin \beta} = \frac{c}{\sin \gamma}$$

-
Unless you know that neither of the unknown angles is going to be obtuse, it's probably better to use the Law of Cosines to find the unknown angles to avoid the ambiguity using a sine to find an angle measure between 0° and 180°. –  Isaac Feb 5 '12 at 21:49