A man starts from a point X and walk 285 m to Y on a bearing of 078\(^o\). He then walks due South to a point Z which is 307 m from X.
(a) Illustrate the information on a diagram.
(b) Find, correct to the nearest whole number, the:
(i) bearing of X from Z;
(ii) distance between Y and Z.
(a) They sketched the required diagram as follows:
(b) (i), using sine rule \(\frac{285}{\sin z} = \frac{307}{\sin 307}\)
Then sin z \(\frac{285 \times \sin 78^o}{307} = \frac{285 \times 0.9781}{307}\) = 0.9080
Z = \(\sin^{-1}(0.9080)\) = 65.23\(^o\)
The bearing of X from Z = 360\(^o\) - 65.23\(^o\) = 294.77\(^o\) \(\approx\) 295\(^o\) or N65\(^o\)W correct to the nearest whole number.
(b)(ii) They obtained < YXZ = 180\(^o\) - 143.23\(^o\) = 36.77\(^o\)
Then, using sine rule, \(\frac{\text{|YZ|}}{\sin 36.77^o} = \frac{307}{\sin 78^o}\) and computing for |YZ|,
|YZ| = \(\frac{307 \times 0.5986}{0.9781}\) = 187.88m
and Therefore, the distance between Y and Z \(\approx\) 188 m correct to the nearest whole number
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