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Trig Help? (1 Viewer)

krakend

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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

dcsd.JPG
 

BLIT2014

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

BLIT2014

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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

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

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