The answer is cooler. Hope this helps.
Answer:
The horizontal component of the truck's velocity is: 23.70 m/s
The vertical component of the truck's velocity is: 3.13 m/s
Explanation:
You have to apply trigonometric identities for a right triangle (because the ramp can be seen as a right triangle where the speed is the hypotenuse), in order to obtain the components of the velocity vector.
The identities are:
Cosα= 
Senα= 
Where H is the hypotenuse, α is the angle, CA is the adjacent cathetus and CO is the opposite cathetus
The horizontal component of the truck's velocity is:
Let Vx represent it.
In this case, CA=Vx, H=24 and α=7.5 degrees
Vx=(24)Cos(7.5)
Vx=23.79 m/s
The vertical component of the truck's velocity is:
Let Vy represent it.
In this case, CO=Vy, H=24 and α=7.5 degrees
Vy=(24)Sen(7.5)
Vy=3.13 m/s
here as it is given that x component of the vector is positive while y component of the vector is negative so we can say the vector must inclined in Fourth quadrant.
So angle must be more than 270 degree and less than 360 degree
Now in order to find the value we can say that




so it is inclined at above angle with X axis in fourth quadrant
Now if angle is to be measured counterclockwise then its magnitude will be

so the correct answer will be 305 degree