A basket of flowers of mass 2kg in a garden. Gravity $g=9.8\frac{m}{s^2}$
The basket is suspended between two tree branches by two ropes. The left rope(S) makes an angle of 54 with the horizontal, and the right rope(T) makes an angle of 43 with the horizontal. Calculate the magnitude in newtons of the tension in each rope.
So far I have
We know that $|W| =2g$,
$$T=|T|\cos(43)i+|T|\sin(43)\hat{j}$$
$$=0.73T\hat{i}+0.68T\hat{j}$$
$$S=-S\cos(54)\hat{i}+S\sin(54)\hat{j}$$
$$=-0.59S\hat{i}+0.81S\hat{j}$$ Also $W=-2g\hat{j}$
$$0.73T\hat{i}+0.68T\hat{j}-0.59S\hat{i}+0.81S\hat{j}-2g\hat{j}=0$$
Resolving this equation in the $\hat{i}$- and $\hat{j}$- directions gives the simultaneous equations
$$0.73T-0.59S=0$$
$$0.68T+0.81S-2g=0$$
I know I have gone wrong somewhere as when I solve these equations both S and T come to 0, really stuck as can't figure out where I have gone wrong!!
Starting from your last system of equations
$$ \begin{cases} 0.73\ T - 0.59\ S = 0 \\ 0.68\ T + 0.81\ S - 2g = 0 \end{cases} $$
I am sure you're familiar with the method of solving couple equations; but let's use the raw way of finding one parameter as a function of the other one and then substitute: in the first one we can find $T$ as
$$T = \frac{0.59}{0.73}S = 0.8082\ S$$
And substitute it into the $T$ in the second equation
$$0.68 \cdot (0.8082\ S) + 0.81\ S - 2g = 0$$
Hence
$$0.549\ S + 0.81\ S = 2g$$
$$1.359\ S = 2g$$
Which means
$$S = \frac{2g}{1.359} = 14.422\ \text{N}$$
And consequently
$$T = 11.656\ \text{N}$$
I used $g = 9.8$ so it's quite a raw result ($g = 9.80665$).