$\delta_A = 0$
$\delta_{triangular\,\,load} - \delta_{fixed\,\,end\,\,moment} - \delta_{reaction\,\,at\,\,A} = 0$
For formulas, see Case 1, Case 4, and Case 5 in superposition method
$\delta_1 - \delta_2 - \delta_3 = 0$
$\dfrac{w_oL^4}{30EI} - \dfrac{M_AL^2}{2EI} - \dfrac{R_AL^3}{3EI} = 0$
$\dfrac{w_oL^2}{30} - \dfrac{M_A}{2} - \dfrac{R_AL}{3} = 0$
$\dfrac{w_oL^2}{30} - \dfrac{M_A}{2} = \dfrac{R_AL}{3}$
$R_A = \dfrac{w_oL}{10} - \dfrac{3M_A}{2L}$
$\theta_A = 0$
$\theta_{triangular\,\,load} - \theta_{fixed\,\,end\,\,moment} - \theta_{reaction\,\,at\,\,A} = 0$
$\theta_1 - \theta_2 - \theta_3 = 0$
$\dfrac{w_oL^3}{24EI} - \dfrac{M_AL}{EI} - \dfrac{R_AL^2}{2EI} = 0$
$\dfrac{w_oL^2}{24} - M_A - \dfrac{R_AL}{2} = 0$
$w_oL^2 - 24M_A - 12R_AL = 0$
$w_oL^2 - 24M_A - 12\left( \dfrac{w_oL}{10} - \dfrac{3M_A}{2L} \right)L = 0$
$w_oL^2 - 24M_A - \frac{6}{5}w_oL^2 + 18M_A = 0$
$-6M_A - \frac{1}{5}w_oL^2 = 0$
$6M_A = -\frac{1}{5}w_oL^2$
$M_A = -\frac{1}{30}w_oL^2$ answer
$R_A = \dfrac{w_oL}{10} - \dfrac{3(-\frac{1}{30}w_oL^2)}{2L}$
$R_A = \dfrac{w_oL}{10} + \dfrac{w_oL}{20}$
$R_A = \frac{3}{20}w_oL$
$M_B = M_A + R_AL - \frac{1}{2}w_oL(\frac{1}{3}L)$
$M_B = -\frac{1}{30}w_oL^2 + (\frac{3}{20}w_oL)L - \frac{1}{2}w_oL(\frac{1}{3}L)$
$M_B = -\frac{1}{30}w_oL^2 + \frac{3}{20}w_oL^2 - \frac{1}{6}w_oL^2$
$M_B = -\frac{1}{20}w_oL^2$ answer
