First, let's draw a diagram representing the distances:
a)
To calculate the straight line distance, let's use the law of cosines, using the interior angle adjacent to the angle 25°, that is, an angle of 155°:
![\begin{gathered} a^2=b^2+c^2-2bc\cdot\cos155°\\ \\ a^2=200^2+280^2-2\cdot200\cdot280\cdot(-0.906)\\ \\ a^2=40000+78400+101472\\ \\ a^2=219872\\ \\ a=468.9\text{ km} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/1gnioqsit8v4ll9n46byqq4l4c4u45nxy9.png)
b)
Now, to find the direction, first let's decompose the 280 vector in horizontal and vertical components:
![\begin{gathered} V_x=280\cdot\cos25°=280\cdot0.906=253.68\\ \\ V_y=280\cdot\sin25°=280\cdot0.4226=118.33 \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/z2bit2du0t69fz6j31y9i0e7uhry6ltyhx.png)
Now, let's add the horizontal component with the 200 vector, then calculate the angle of the resulting vector:
![\begin{gathered} V_x=253.68+200=453.68\\ \\ \theta=\tan^(-1)((V_y)/(V_x))=\tan^(-1)((118.33)/(453.68))=14.62° \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/jtb4dv4ipwg1blzp4mxwk2214p9t3f91br.png)
c)
The answer is approximately correct because the sine and cosine values used to calculate are not exact values, they are rounded values. Therefore the final result will not be 100% accurate.