SOLUTION:
Case: Cosine rule:
Given:
Method:
Using cosine rule:
[tex]\begin{gathered} c^2=a^2+b^2-2bc\cos C \\ c^2=10^2+14^2-2(10)(14)\cos135\degree \\ c^2=100+196-280(-0.7071) \\ c^2=296+197.99 \\ c^2=493.99 \\ c=\sqrt{493.99} \\ c=22.226 \end{gathered}[/tex]
Next, we find the bearing of B from A.
This is the angle at A
Using sine Rule:
[tex]\begin{gathered} \frac{\sin A}{a}=\frac{\sin C}{c} \\ \frac{\sin A}{10}=\frac{\sin135}{22.226} \\ \sin A=\frac{10\times\sin135}{22.226} \\ \sin A=\frac{10\times0.7071}{22.226} \\ \sin A=0.3181 \\ A=\sin^{-1}0.3181 \\ A=18.55\degree \end{gathered}[/tex]
Final answer: To 3 and 2 decimal places
A) The distance between A and B is 22.226 hm
B) The bearing of B from A is 18.55 degrees
