Q22Basic Civil Engineering

In a closed traverse surveying loop, verifying the geometric accuracy of the measured bearings is critical. We are given the Fore Bearings (FB) of a five-sided polygon traversed in an anti-clockwise direction. To find the interior angles, we must first calculate the Back Bearings (BB) for each line. The fundamental relationship is: $\text{BB}=\text{FB}\pm 180^{\circ }$ (Use $+$ if $FB<180^{\circ }$, use $-$ if $FB>180^{\circ }$).

Step 1: Calculate Back Bearings - Line AB: $FB=105^{\circ }15^{\prime }$. Since $FB<180^{\circ }$, $BB\text{ of }AB=105^{\circ }15^{\prime }+180^{\circ }=285^{\circ }15^{\prime }$ - Line BC: $FB=20^{\circ }00^{\prime }$. Since $FB<180^{\circ }$, $BB\text{ of }BC=20^{\circ }00^{\prime }+180^{\circ }=200^{\circ }00^{\prime }$ - Line CD: $FB=316^{\circ }30^{\prime }$. Since $FB>180^{\circ }$, $BB\text{ of }CD=316^{\circ }30^{\prime }-180^{\circ }=136^{\circ }30^{\prime }$ - Line DE: $FB=187^{\circ }15^{\prime }$. Since $FB>180^{\circ }$, $BB\text{ of }DE=187^{\circ }15^{\prime }-180^{\circ }=07^{\circ }15^{\prime }$ - Line EA: $FB=122^{\circ }45^{\prime }$. Since $FB<180^{\circ }$, $BB\text{ of }EA=122^{\circ }45^{\prime }+180^{\circ }=302^{\circ }45^{\prime }$

Step 2: Calculate Interior Angles For an anti-clockwise closed traverse, the interior angle at any station can be found by subtracting the Fore Bearing of the forward line from the Back Bearing of the preceding line. If the result is negative, add $360^{\circ }$. $\text{Interior Angle at Station}=\text{BB of preceding line}-\text{FB of forward line}$

• Angle A: At station A, preceding line is EA, forward is AB. $\angle A=BB\text{ of }EA-FB\text{ of }AB=302^{\circ }45^{\prime }-105^{\circ }15^{\prime }=197^{\circ }30^{\prime }$ Wait, let's draw it. For anti-clockwise traversing, the interior angle is usually calculated as $BB-FB$. But let's verify. Let's use a universal formula: Included Angle = $FB\text{ of next line}-BB\text{ of previous line}$. If negative, add $360^{\circ }$. This gives the clockwise angle. For anti-clockwise traverse, interior angle = $360^{\circ }-\text{clockwise angle}$, or simply $BB\text{ of previous}-FB\text{ of next}$. Let's re-calculate using $\angle =B{B}_{prev}-F{B}_{next}$: - Angle A: $302^{\circ }45^{\prime }(EA)-105^{\circ }15^{\prime }(AB)=197^{\circ }30^{\prime }$. Since it's > 180, this is the EXTERIOR angle. Interior $\angle A=360^{\circ }-197^{\circ }30^{\prime }=162^{\circ }30^{\prime }$.
• Angle B: Preceding AB, forward BC. $BB\text{ of }AB=285^{\circ }15^{\prime }$. $FB\text{ of }BC=20^{\circ }00^{\prime }$. Angle $=285^{\circ }15^{\prime }-20^{\circ }00^{\prime }=265^{\circ }15^{\prime }$ (Exterior). Interior $\angle B=360^{\circ }-265^{\circ }15^{\prime }=94^{\circ }45^{\prime }$.
• Angle C: Preceding BC, forward CD. $BB\text{ of }BC=200^{\circ }00^{\prime }$. $FB\text{ of }CD=316^{\circ }30^{\prime }$. Angle $=200^{\circ }00^{\prime }-316^{\circ }30^{\prime }=-116^{\circ }30^{\prime }$. Since negative, add $360^{\circ }$: $-116^{\circ }30^{\prime }+360^{\circ }=243^{\circ }30^{\prime }$ (Exterior). Interior $\angle C=360^{\circ }-243^{\circ }30^{\prime }=116^{\circ }30^{\prime }$.
• Angle D: Preceding CD, forward DE. $BB\text{ of }CD=136^{\circ }30^{\prime }$. $FB\text{ of }DE=187^{\circ }15^{\prime }$. Angle $=136^{\circ }30^{\prime }-187^{\circ }15^{\prime }=-50^{\circ }45^{\prime }$. Add $360^{\circ }$: $309^{\circ }15^{\prime }$ (Exterior). Interior $\angle D=360^{\circ }-309^{\circ }15^{\prime }=50^{\circ }45^{\prime }$.
• Angle E: Preceding DE, forward EA. $BB\text{ of }DE=07^{\circ }15^{\prime }$. $FB\text{ of }EA=122^{\circ }45^{\prime }$. Angle $=07^{\circ }15^{\prime }-122^{\circ }45^{\prime }=-115^{\circ }30^{\prime }$. Add $360^{\circ }$: $244^{\circ }30^{\prime }$ (Exterior). Interior $\angle E=360^{\circ }-244^{\circ }30^{\prime }=115^{\circ }30^{\prime }$.

Step 3: Apply Geometric Check The sum of the interior angles of any closed polygon with '$n$' sides is given by the formula $(2n-4)\times 90^{\circ }$. For a 5-sided traverse (pentagon), $n=5$. Theoretical Sum = $(2\times 5-4)\times 90^{\circ }=6\times 90^{\circ }=540^{\circ }$. Let's sum our calculated interior angles: Sum $=\angle A+\angle B+\angle C+\angle D+\angle E$ Sum $=162^{\circ }30^{\prime }+94^{\circ }45^{\prime }+116^{\circ }30^{\prime }+50^{\circ }45^{\prime }+115^{\circ }30^{\prime }$ $=162.5+94.75+116.5+50.75+115.5=540.0^{\circ }$ Sum = $540^{\circ }00^{\prime }$. Since the calculated sum exactly matches the theoretical sum, the interior angles are correct and the traverse mathematically closes without angular error.