Answer:
[tex]\vec{v}= <9\cos 10^{\circ}, 9\sin 10^{\circ}>\text{ or } \approx <8.86326977711,1.562833599 >[/tex]
Step-by-step explanation:
Component form of a vector is given by [tex]\vec{v}=<i, j>[/tex], where [tex]i[/tex] represents change in x-value and [tex]j[/tex] represents change in y-value. The magnitude of a vector is correlated the Pythagorean Theorem. For vector [tex]\vec{v}=<i,j>[/tex], the magnitude is [tex]||v||=\sqrt{i^2+j^2[/tex].
190 degrees counterclockwise from the positive x-axis is 10 degrees below the negative x-axis. We can then draw a right triangle 10 degrees below the horizontal with one leg being [tex]i[/tex], one leg being [tex]j[/tex], and the hypotenuse of the triangle being the magnitude of the vector, which is given as 9.
In any right triangle, the sine/sin of an angle is equal to its opposite side divided by the hypotenuse, or longest side, of the triangle.
Therefore, we have:
[tex]\sin 10^{\circ}=\frac{j}{9},\\j=9\sin 10^{\circ}[/tex]
To find the other leg, [tex]i[/tex], we can also use basic trigonometry for a right triangle. In right triangles only, the cosine/cos of an angle is equal to its adjacent side divided by the hypotenuse of the triangle. We get:
[tex]\cos 10^{\circ}=\frac{i}{9},\\i=9\cos 10^{\circ}[/tex]
Verify that [tex](9\sin 10^{\circ})^2+(9\cos 10^{\circ})^2=9^2\:\checkmark[/tex]
Therefore, the component form of this vector is [tex]\vec{v}=\boxed{<9\cos 10^{\circ}, 9\sin 10^{\circ}>}\approx \boxed{<8.86326977711,1.562833599 >}[/tex]
