2007 WAEC Mathematics Theory Y is 60 km away from X on a bearing of 135°. Z is 80...

Y is 60 km away from X on a bearing of 135°. Z is 80 km away from X on a bearing of 225°. Find the :

(a) distance of Z from Y ;

(b) bearing of Z from Y.

(a) < ZXY = 225° - 135°

= 90°.

\(\therefore /ZY/ = 80^{2} + 60^{2}\)

= \(6400 + 3600 = 10000\)

\(/ZY/ = \sqrt{10000}\)

= 100 km.

(b) From \(\Delta\) XYZ,

\(\tan \theta = \frac{opp}{adj}\)

\(\tan \theta = \frac{80}{60} = 1.333\)

\(\theta = \tan^{-1} (1.333)\)

\(\theta = 53.13°\)

Bearing of Z from Y = \(360° - (45° + 53.13°)\)

= \(360° - 98.13°\)

= \(261.87° \approxeq 262°\) (3 significant figures)

{{ settings.no_comment_msg ? settings.no_comment_msg : 'There are no comments' }}