Theodolites Compass And Traverse Surveying Practice Questions

Let, length of PQ and QR are $L_{1}$ and $L_{2}$ respectively. From P, coordiantes of $Q =(100+L_{1}\cos 30^{\circ},200 \left.+L_{1} \sin 30^{\circ}\right)$ From R, coordinates of $Q=\left(1000-L_{2},\cos 45^{\circ},\right. \left.1000-L_{9} \sin 45^{\circ}\right)$ But, $100 + L_{1} \cos 30^{\circ} = 100-L_{2}\cos45^{\circ}$ $\Rightarrow \frac{\sqrt{3}}{2} L_{1}+\frac{1}{\sqrt{2}} L_{9}=900 \;\;\;...(i)$ and $200+L_{1} \sin 30^{n}=1000-L_{2} \sin 46^{\prime}$ $\Rightarrow \frac{1}{2} L_{1}+\frac{1}{\sqrt{2}} L_{9}=800 \;\;\; ...(ii)$ Solving (i) and (ii) we get $L_{1}=273.206 \mathrm{m}\text{ and }\mathrm{L}_{2}=936.166 \mathrm{m}$