SOLUTION:
Case: Cosine rule:
Given:
Method:
Using cosine rule:
![\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=√(493.99) \\ c=22.226 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/40r4rl929rb4txzc99yci29eyzsehnumlv.png)
Next, we find the bearing of B from A.
This is the angle at A
Using sine Rule:
![\begin{gathered} (\sin A)/(a)=(\sin C)/(c) \\ (\sin A)/(10)=(\sin135)/(22.226) \\ \sin A=(10*\sin135)/(22.226) \\ \sin A=(10*0.7071)/(22.226) \\ \sin A=0.3181 \\ A=\sin^(-1)0.3181 \\ A=18.55\degree \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/e3irafviekv50d0o00x3jbgxzkskn0pq8d.png)
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