Answer
Magnitude of $\vec{T}_R=435.8N$ and it is directed at angle of $5.64^o$ to the horizontal.
Work Step by Step
Take rightward and upward to be our $+i$ and $+j$ directions, respectively. We rewrite all the forces in unit-vector notation:
1) Weight: $m\vec{g}=(-151N)j$
2) Left tension: $\vec{T}_L=-(447\cos14N)i+(447\sin14N)j$ $$\vec{T}_L=(-433.72N)i+(108.14N)j$$
3) Right tension: $\vec{T}_R=ai+bj$
As there are no more forces, we have $$\sum \vec{F}=m\vec{g}+\vec{T}_L+\vec{T}_R$$ $$\sum \vec{F}=(a-433.72N)i+(b-42.86N)j$$
Since the limb is static, there is no acceleration whatsoever, which means $\sum \vec{F}=0$
Therefore, $a-433.72N=0$ and $a=433.72N$
$b-42.86N=0$ and $b=42.86N$
- Magnitude of $\vec{T}_R=\sqrt{433.72^2+42.86^2}=435.8N$
- Direction: let's call $\theta$ the angle relative to the horizontal, we have $\tan\theta=\frac{42.86}{433.72}=0.0988$ so $\theta=5.64^o$
