Let’s solve Question 9 clean and fast.

Given (from the problem statement):

• Latitudes failed to close by +0.30 ft
  
  

• Departures failed to close by +0.45 ft
  
  

To close the traverse, the closing line must be:

• ΔN = −0.30 ft
  
  

• ΔE = −0.45 ft
  
  

So the closing line lies in the SW quadrant.

Compute bearing angle:

tan⁡θ=∣ΔE∣∣ΔN∣=0.450.30=1.5\tan \theta = \frac{|\Delta E|}{|\Delta N|} = \frac{0.45}{0.30} = 1.5tanθ=∣ΔN∣∣ΔE∣​=0.300.45​=1.5 θ=tan⁡−1(1.5)≈56.31∘≈56∘18′\theta = \tan^{-1}(1.5) \approx 56.31^\circ \approx 56^\circ 18'θ=tan−1(1.5)≈56.31∘≈56∘18′

Bearing:
S 56° 18′ W

✅ Correct answer: B

سؤال اصلاح‌شده (طوری که C درست شود)

A closed traverse has the following misclosure:

• Latitudes failed to close by −0.30 ft
  
  

• Departures failed to close by −0.45 ft
  
  

The sum of the horizontal traverse distances is 4,950.34 ft.

Question:
What is the bearing of the closing line?

A) S 56°18′ W
B) S 56°18′ E
C) N 56°18′ E
D) N 56°18′ W

چرا این‌بار C درست است؟

• Misclosure:
  
  

  • ΔN = −0.30 → جنوب
    
    

  • ΔE = −0.45 → غرب
    
    

• Closing line = برعکس misclosure:
  
  

  • ΔN = +0.30 → شمال
    
    

  • ΔE = +0.45 → شرق
    
    

→ ربع NE

زاویه:

tan⁡θ=0.450.30=1.5⇒θ≈56∘18′\tan \theta = \frac{0.45}{0.30} = 1.5 \Rightarrow \theta \approx 56^\circ 18'tanθ=0.300.45​=1.5⇒θ≈56∘18′

✅ N 56°18′ E (گزینه C)