Question

The lengths and bearings of a closed traverse PQRSP are given below.

Line	Length (m)	Bearing (WCB)
PQ	200	$0^{\circ }$
QR	1000	${45}^{\circ }$
RS	907	${180}^{\circ }$
SP	?	?

The missing length and bearing, respectively of the line SP are

• A. $907\text{m and}{270}^{\circ }$
• B. $707\text{m and}{270}^{\circ }$
• C. $207\text{m and}{270}^{\circ }$
• D. $707\text{m and}{180}^{\circ }$

Answer 1

The correct option is B $707\text{m and}{270}^{\circ }$

Missing length of line,
SP = L

WCB of line,
$SP=\theta$

In a closed traverse,

Sum of latitudes i.e.,
$\sum L\cos \theta =0$

$200\cos 0^{\circ }+1000\cos {45}^{\circ }+907\cos {180}^{\circ }+L\cos \theta =0$

$\Rightarrow L\cos \theta =-0.107...\left(1\right)$

Sum of Departures i.e.,
$\sum L\sin \theta =0$

$200\sin 0^{\circ }+1000\sin {45}^{\circ }+907\sin {180}^{\circ }+L\sin \theta =0$

$=L\sin \theta =-707.107...\left(2\right)$

Dividing equation (2) by equation (1)
$\tan \theta =6608.475$

$\Rightarrow \theta ={89.99}^{\circ }$

$\simeq {90}^{\circ }$

Since both $L\cos \theta andL\sin \theta$ have negative value hence third quadrant.

$So,\theta =90+180={270}^{\circ }$

$andL=\frac{-707.107}{\sin {270}^{\circ }}=707.1m$