Question

Hi:) how to do this?

Answer 1

b) 15.2km

c) Bearing= 263.8° (1 d.p.),

AD= 25.5 km (3 s.f.)

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,

`(sin∠ABD)/(AD) = (sin∠ADB)/(AB) \\ (sin65°)/(AD) = (sin83.829°)/(28) \\ AD(sin83.829°) = 28(sin65)° \\ AD = (28(sin65°))/(sin83.829°) \\ AD = 25.5km \: (3 \: s.f.)`

Feel free to ask if you have any queries :)