- Thread starter krakend
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(b)

(i) PAQ =135°.

(ii) 185°.

(iii) PAB 73°

(i) PAQ =135°.

(ii) 185°.

(iii) PAB 73°

(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)

Nah its fine, good luck for general maffs.Thank you so much! Sorry for the last minute question haha

https://thsconline.github.io/s/?view=5310&id=0ByEFYhkkDQBKaFNNVWpET245YWc&n=2004 SolutionsThank you so much! Sorry for the last minute question haha

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