Solution
The graph above is the graphical representation of the airplane movement
For angle A
[tex]A=90+45=135^{\circ}[/tex]Note: Cosine Rule Formula
Usimg the cosine rule, we have
[tex]\begin{gathered} a^2=b^2+c^2-2bccosC \\ a^2=425^2+100^2-2(425)(100)cos(135) \\ a^2=250729.0764 \\ a=\sqrt{250729.0764} \\ a=500.7285456 \\ a=500.73kmh^{-1} \end{gathered}[/tex]Therefore, the real speed is
[tex]\begin{equation*} 500.73kmh^{-1} \end{equation*}[/tex]We are left with getting the real trajectory
The real trajectory is angle C
[tex]C=45+\theta[/tex]We will find theta by using the sine rule
[tex]\begin{gathered} \frac{a}{sinA}=\frac{c}{sin\theta} \\ \frac{500.7285456}{sin135}=\frac{100}{sin\theta} \\ sin\theta=\frac{100sin135}{500.7285456} \\ sin\theta=0.1412155922 \\ \theta=8.11819343 \\ \theta=8.118^{\circ} \end{gathered}[/tex]Thus,
[tex]\begin{gathered} C=45+8.118 \\ C=53.118^{\circ} \end{gathered}[/tex]Therefore, the answer is
[tex]\begin{gathered} Real\text{ }Trajectory=53.118^{\circ} \\ Real\text{ }Speed=500.73kmh^{-1} \end{gathered}[/tex]
