Angles in a static equilibrium

This is a typical solution trick when you have a system involving sine and cosine of the same unknown angle: Re-organize the equations so that you have $$cosalpha = stuff $$ and $$sinalpha = other stuff.$$ Square these equations and add them. The angle $alpha$ is eliminated because $$sin^2alpha + cos^2alpha = 1.$$

In your case, it's nice that the coefficiencts on both $alpha$ terms are the same, and also on both $beta$ terms. You can use a double angle formula for the remaining $beta$ terms to solve for $beta.$ Then you can solve for $alpha$. Remember that tool, and teach it to someone else.

You can eliminate the angle $alpha$ from the equations with the trick the other answers give you ^*. But then you will end up with an equation of the form

$$ A cos beta + B sin beta + C = 0$$

To solve this do the following transformation

$$ left. begin{align} A & = R cos psi B & = R sin psi end{align} right} begin{aligned} R & = sqrt{A^2+B^2} psi & = arctanleft( frac{B}{A} right) end{aligned} $$

The equation is now $$ cosbetacospsi + sinbeta sinpsi = cos(beta-psi) = -frac{C}{R} $$

which is solved for

$$ begin{split} beta & = arccosleft( -frac{C}{R} right) + psi & = arccosleft( -frac{C}{sqrt{A^2+B^2}} right) + arctanleft( frac{B}{A} right)end{split}$$

_footnotes:

• make the equations of this form $$begin{align} cos alpha & = a cosbeta+c_x sin alpha & = -a sin beta + c_y end{align}$$
• square both sides and add them for $$ 1 = 2 a c_x cosbeta - 2 a c_y sinbeta + c_x^2 + c_y^2 +a^2 $$ $$ left(2 a c_xright) cosbeta + left(- 2 a c_yright) sinbeta + left(c_x^2 + c_y^2 +a^2-1right) = 0 $$
• Match the $A$, $B$ and $C$ coefficients.
• Once $beta$ is known, then divide the two equations above for $$ tan alpha = frac{c_y - a sinbeta}{c_x + a cosbeta} $$

Edit 1

Here is the actual solution:

$$left. begin{align} -313.9 cos(alpha) + 619 cos(beta) + 60 & = 0 313.9 sin(alpha) + 619 sin(beta) - 882.9 & = 0 end{align} right} begin{aligned} 313.9 cos(alpha) & = 619 cos(beta) + 60 313.9 sin(alpha) & = - 619 sin(beta) + 882.9 end{aligned} $$

Square and add the two equations (on each side) to get

$$ left. 98533.21 = 74280 cos(beta) - 1093030.2 sin(beta) + 1166273.41 right} 74280 cos(beta) - 1093030.2 sin(beta) + 1067740.2 = 0 $$

$$ begin{aligned} beta & = arccosleft( -frac{C}{sqrt{A^2+B^2}} right) + arctanleft( frac{B}{A} right) A & = 74280 B & = -1093030.2 C & = 1067740.2 beta &= 1.41284652 = 80.9501426° end{aligned} $$

Finally, $alpha$ can be solved with the 2nd equation:

$$ sin(alpha) = 2.81267919-1.97196559 sin(beta) $$ $$ alpha = 1.04567064 = 59.9125144° $$

Now you can plug the values of $alpha$ and $beta$ into the two original equations to confirm it balances the forces.