A mass M of 100 kg is suspended with the use of strings A, B and C as shown in fig. Here W is 

vertical wall and of is rigid horizontal rod. The tension in string B is

Let T be the tension in the string C. Then

$\mathrm{T} \cos ⁡{45}^{∘}=\mathrm{Mg}$   â€¦(1)

and $\mathrm{T} \sin ⁡{45}^{∘}=\text{ tension in }\mathrm{B}$   â€¦(2)

$\text{From eq. (i), }\mathrm{T}=\frac{\mathrm{Mg}}{\cos ⁡{45}^{∘}}=\sqrt{2}\mathrm{Mg}$

From eq. (ii), tension in B

$\begin{array}{l}=\mathrm{Tsin}⁡{45}^{∘}=\left(\sqrt{2}\mathrm{Mg}\right)×(1/\sqrt{2})\\ =\mathrm{Mg}=100\mathrm{g}\text{ newton. }\end{array}$