Answer:
b) 15.2km
c) Bearing= 263.8° (1 d.p.),
AD= 25.5 km (3 s.f.)
Step-by-step explanation:
Please see attached picture for full solution.
In picture 3, you could also use trigonometry to find AW, just like how we did to find W.
Picture 4:
WD= 16 -15.250
WD= 0.75
Picture 5:
Bearing of D from A
= reflex ∠N₂AD
= 295° - ∠BAD
∠BAD
= ∠BAW -∠WAD
= 33° -1.829°
= 31.171°
Thus, bearing of D from A
= 295° - 31.171°
= 263.8° (1 d.p.)
c) Distance of D from A= length AD
Look at triangle BAD:
∠DBA= 65°
∠ADB= 180° -65° -31.171° (∠ sum of △ADB)
∠ADB= 83.829°
Using sine rule,
[tex] \frac{sin∠ABD}{AD} = \frac{sin∠ADB}{AB} \\ \frac{sin65°}{AD} = \frac{sin83.829°}{28} \\ AD(sin83.829°) = 28(sin65)° \\ AD = \frac{28(sin65°)}{sin83.829°} \\ AD = 25.5km \: (3 \: s.f.)[/tex]
Feel free to ask if you have any queries :)
