91Ó°ÊÓ

Strike-slip $AC=22.0m$and dip-slip$AD=17.0\mathrm{m}$

The inclination angle of the fault plane to the horizontal is$\mathrm{ϕ}=52°$

The vector addition theorem can be used to find the net displacement and its vertical component can be found by using trigonometry and the information given in the figure.

Formula:

$\stackrel{→}{AD}=\stackrel{→}{AD}+\stackrel{→}{AD}………\left(\mathrm{i}\right)$

$s\mathrm{in}\mathrm{ϕ}=\frac{\mathrm{opposite}\mathrm{side}}{\mathrm{hypotenuse}}……….\left(\mathrm{ii}\right)$

$\stackrel{→}{AB}$and$\stackrel{→}{AD}$ are the horizontal and vertical component of$\stackrel{→}{AB}$

According to the triangle vector addition law,

$\stackrel{→}{AB}=\stackrel{→}{AC}+\stackrel{→}{AD}$

$$\begin{gathered} A{B}^2=A{C}^2+A{D}^2 \\ AB=\sqrt{A{C}^2+A{D}^2} \\ =\sqrt{{22}^3+{17}^2} \\ =27.8\mathrm{m} \end{gathered}$$

Hence, the magnitude of net displacement is 27.8 m

Now the fault is inclined at an angle$\mathrm{ϕ}$ to the horizontal, hence, the vertical component of$\stackrel{→}{AB}$ is the opposite side of that angle$\mathrm{ϕ}$

The vertical component of $\stackrel{→}{AB}$,

$$\begin{gathered} A{B}_v=AD\sin \mathrm{Ï•} \\ =17\sin 52 \\ =13.4\mathrm{m} \end{gathered}$$

Therefore, the vertical component of$\stackrel{→}{AB}=13.4\mathrm{m}$ .