Dear experts kindly answer the following question

Solution:
Since, the ladder is at rest so, net force on it must be zero. As well as it must be in rotational equilibrium.


Let length of the ladder be l.
Since, the ladder is at rest,
$$\begin{gathered} N-Mg=0 ...\left(1\right) \\ N\text{'}-f_s=0 ...\left(2\right) \\ \mathrm{So}, N=Mg=10\times 10=\mathbf{100}\mathbf{ }\mathbf{N} \\ \mathrm{The} \mathrm{ladder} \mathrm{must} \mathrm{be} \mathrm{in} \mathrm{rotational} \mathrm{equilibrium}. \\ \mathrm{So} \mathrm{taking} \mathrm{moment} \mathrm{of} \mathrm{force} \mathrm{about} \mathrm{point} \mathrm{C}, \\ N\times l\cos 37°-f_{s}l\cos 53°-Mg\times \frac{l}{2}\cos 37°-N\text{'}\times 0=0 \\ \Rightarrow f_s=\frac{N\times l\cos 37°-Mg\times \frac{l}{2}\cos 37°}{l\cos 53°} \\ =\frac{N\cos 37°-{\displaystyle \frac{Mg\cos 37°}{2}}}{\cos 53°} \\ \left(N=100 \mathrm{N}, Mg\mathit{ }= 100 \mathrm{N}, \cos 37°=\frac{4}{5} \mathrm{and} \cos 53°=\frac{3}{5}\right) \\ \Rightarrow f_s=66.66 \mathrm{N} \end{gathered}$$
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So, N = 100 N and f_s = 66.66 N
Hence, the correct answer is option (a).