Need help with a trigonometry problem (w/ picture) I have this trigonometry problem I got when programming a code library for cameras in games. I made a picture in Paint to explain the problem as simple as possible. Here's a link:

The known values are random but it shouldn't be a problem in this case. However, I want to know how, and if, I can get the unkown values in the picture.
Thanks in advance!
 A: Yes it can be gotten
$$
A=x,
B=50-x,
C=80-x,
D=50+x,
$$
$$ \frac{3}{sin(x)}=\frac{b}{sin(80-x)}$$
$$ \frac{5}{sin(50-x)}=\frac{b}{sin(50+x)}$$
$$ \frac{3*sin(80-x)}{sin(x)}=\frac{5*sin(50+x)}{sin(50-x)}$$
$$ 3*sin(80-x)*sin(50-x)=5*sin(50+x)*sin(x)$$
$$ 3/2*(-cos(130-2x)+cos(30))=5/2*(-cos(50+2x)+sin(50))$$
$$  130-2x=k $$
$$cos(180-\theta)=-cos(\theta)$$
$$ 3/2*(-cos(k)+cos(30))=5/2*(cos(k)+sin(50))$$
$$4*cos(k)=3/2*cos(30)-5/2*sin(50) $$
A: Hint...you can apply the cotangent rule for triangles and obtain $$8\cot80=3\cot A-5\cot(50-A)$$ 
Then you can form and solve a quadratic equation for $\tan A$, which is fairly straightforward, and thus obtain all the unknown quantities after application of the Sine rule in the smaller triangles.
A: Let $A_1$ be the Area of the Traingle having sides a,b,3;
Let $A_2$ be the Area of triangle having sides b,c,5;
And A be the area of wholes triangle;
(i.e)$A_1+A_2=A $
$$ A_1=(0.5)3b\sin(100);A_2=(0.5)5b\sin(80);A=(0.5)ac\sin(50)$$
$$ 3b\sin(100)+5b\sin(80)=ac\sin(50)$$
$$ ac=\frac{3b\sin(100)+5b\sin(80)}{\sin(50)}-.......EQN1$$
$$ \text{Now Applying Cosine law in triangle 1}:a^2=b^2+9-6b\cos(100)$$
$$ \text{now Applying for Secon one}:c^2=b^2+25-10b\cos(80)$$
$$\text{Multiplying both}:a^2c^2=(b^2+9-6b\cos(100))(b^2+25-10b\cos(80))$$
$$ac=\sqrt{(b^2+9-6b\cos(100))(b^2+25-10b\cos(80))}$$
$$ \text{by equating both equation we can find b and automatically we find unknowns}$$
A: 
The law of cosines:
$$
\begin{cases}
\text{a}^2=\text{b}^2+\text{c}^2-2\text{b}\text{c}\cos\left(\alpha\right)\\
\text{b}^2=\text{a}^2+\text{c}^2-2\text{a}\text{c}\cos\left(\beta\right)\\
\text{c}^2=\text{a}^2+\text{b}^2-2\text{a}\text{b}\cos\left(\gamma\right)
\end{cases}
$$
And the law of sines:
$$\frac{\text{a}}{\sin\left(\alpha\right)}=\frac{\text{b}}{\sin\left(\beta\right)}=\frac{\text{c}}{\sin\left(\gamma\right)}$$
Where:



So, we can set up the equations for your problem:
$$\begin{cases}
\text{a}^2=\text{b}^2+8^2-2\text{b}\cdot8\cdot\cos\left(\alpha\right)\\
\text{b}^2=\text{a}^2+8^2-2\text{a}\cdot8\cdot\cos\left(\beta\right)\\
8^2=\text{a}^2+\text{b}^2-2\text{a}\text{b}\cos\left(50^\circ\right)\\
\frac{\text{a}}{\sin\left(\alpha\right)}=\frac{\text{b}}{\sin\left(\beta\right)}=\frac{8}{\sin\left(50^\circ\right)}\\
\alpha+\beta=130^\circ
\end{cases}\Longleftrightarrow
\begin{cases}
\text{a}^2=\text{b}^2+64-16\text{b}\cos\left(130^\circ-\beta\right)\\
\text{b}^2=\text{a}^2+64-16\text{a}\cos\left(\beta\right)\\
64=\text{a}^2+\text{b}^2-2\text{a}\text{b}\cos\left(50^\circ\right)\\
\sin\left(\beta\right)=\frac{\text{b}\sin\left(50^\circ\right)}{8}\\
\alpha=130^\circ-\beta
\end{cases}
$$
So, we get:
$$128-16\cdot\frac{8\sin\left(\beta\right)}{\sin\left(50^\circ\right)}\cdot\cos\left(130^\circ-\beta\right)-16\text{a}\cos\left(\beta\right)=0$$
