Trig Help? (1 Viewer)

[krakend]

Hiya! Just having trouble with the first part of the trig question. It's from 2004 (Question 24b) if you're interested. I can't find an explanation or answer anywhere Thank youu

 

[BLIT2014]

(b)
(i)
PAQ =135°.
(ii) 185°.
(iii) PAB
73°

 

[BLIT2014]

B)

(i) Let x lie due east of A (45 degrees) -> East (Sorry can't be bothered drawing the diagram)

<PAQ =<PAX+<XAQ
=90° +45 °
=135°

ii) Bearing is 135°+50° =185°

iii) Use the cosine rule

Cos <PAB =(p^2+b^2-a^2)/(2pb)
= (31^2+28^2-35^2)/(2*31*28)
= 0.299539...

Therefore, <PAB =72.57...
≈73° (nearest degree)

 

[krakend]

Thank you so much! Sorry for the last minute question haha

 

[BLIT2014]

Thank you so much! Sorry for the last minute question haha

Nah its fine, good luck for general maffs.

 

[BLIT2014]

Thank you so much! Sorry for the last minute question haha

https://thsconline.github.io/s/?view=5310&id=0ByEFYhkkDQBKaFNNVWpET245YWc&n=2004 Solutions

Wait I've found the answers on dan964 website so you can see a diagram drawn.