91Ó°ÊÓ

The equation of first wave is$y_1=10\sin \mathrm{Ó\neg }t$

The equation of second wave is $y_2=15\sin \left(\mathrm{Ó\neg }t+30°\right)$

The equation of third wave is $y_3=5\sin \left(\mathrm{Ó\neg }t-45°\right)$

The amplitude of the resultant wave is equal to the vector sum of the amplitude of each wave.

The horizontal component of the resultant wave is given as:

$$\begin{gathered} y_h=10\cos 0°+15\cos 30°+5c\mathrm{os}\left(-45°\right) \\ {y}_h=10+13+3.54 \\ {y}_h=26.54 \end{gathered}$$

The vertical component of the resultant wave is given as:

$$\begin{gathered} y_v=10\sin 0°+15\sin 30°+5\sin \left(-45°\right) \\ {y}_v=3.96 \end{gathered}$$

The resultant amplitude of waves is given as:

$$\begin{gathered} y_r=\sqrt{y_h^2+y_v^2} \\ {y}_r=\sqrt{{\left(26.54\right)}^2+{\left(3.96\right)}^2} \\ {y}_r=26.83 \end{gathered}$$

The direction of the resultant wave is given as:

$\begin{array}{rcl}\tan \mathrm{θ}& =& \frac{y_v}{y_h}\\ \tan \mathrm{θ}& =& \frac{3.96}{26.54}\\ \mathrm{θ}& =& 8.5°\end{array}$

The sum of the wave is given as:

$y=y_r\sin \left(\mathrm{Ó\neg }t+\mathrm{θ}\right)$

Substitute all the values in equation.

$y=\left(26.83\right)\sin \left(\mathrm{Ó\neg }t+8.5°\right)$

Therefore, the sum of wave is $\left(26.83\right)\sin \left(\mathrm{Ó\neg }t+8.5°\right)$.