Question

The length and bearing of a closed traverse ABCDA are given

Line	Length (m)	Bearing (WCB)
AB	300	$0^{\circ }$
BC	1200	${45}^{\circ }$
CD	800	${180}^{\circ }$
DA	?	?

The missing length and bearing of line DA are

• A. $938and{258}^{\circ }$
• B. $928and{228}^{\circ }$
• C. $918and{248}^{\circ }$
• D. $928and{248}^{\circ }$

Answer 1

The correct option is C $918and{248}^{\circ }$

Let, missing length of line DA = L

WCB of line DA = $\theta$

In a closed traverse,
$\sum L=0$
i.e. sum of latitudes = 0

$\Rightarrow 300\cos 0^{\circ }+1200\cos {45}^{\circ }+800\cos {180}^{\circ }+L\cos \theta =0$

$\therefore L\cos \theta =-348.53...\left(i\right)$

Now,
$\sum D=0$
i.e. sum of departure = 0

$\Rightarrow 300\sin 0^{\circ }+1200\sin {45}^{\circ }+800\sin {180}^{\circ }+L\sin \theta =0$
$L\sin \theta =-848.53...\left(ii\right)$

From (i) and (ii),
$tan\theta =\frac{848.53}{348.53}=2.43$

$\theta ={67.67}^{\circ }$

Since both $L\cos \theta$ and $L\sin \theta$ have negative value. Hence, $\theta$ lie in third quadrant.

$\theta ={67.67}^{\circ }+{180}^{\circ }$

$\theta ={247.67}^{\circ }$

And length of line DA = $\frac{-848.53}{sin{247.67}^{\circ }}=917.32m$