Answer :
Given,
The distances; a=30 m
b=60 m
Angles; θ=E25°S
α=E40°N
From the diagram, ∠A is given by,
[tex]\angle A=180\degree-\theta-\alpha[/tex]On substituting the known values,
[tex]\begin{gathered} \angle A=180\degree-25\degree-40\degree \\ =115\degree^{} \end{gathered}[/tex]From the cos rule,
[tex]d^2=a^2+b^2-2ab\cos A[/tex]On substituting the known values,
[tex]\begin{gathered} d^2=30^2+60^2-2\times30\times60\times\cos 115\degree \\ \Rightarrow d=\sqrt[]{30^2+60^2-2\times30\times60\times\cos 115\degree} \\ =77.6\text{ m} \end{gathered}[/tex]Thus the total displacement of the boy is 77.6 m
