Question

a vector is in standard position, with its terminal point in the second quadrant and an x - coordinate of - 5. the vector has a magnitude of $sqrt{61}$. complete the statements describing the vector. the y - coordinate of the vector to the nearest tenth is. the direction angle to the nearest whole number is.

Explanation:

Step1: Find the $y$-coordinate

Let the vector be $\vec{v}=(x,y)$ with $x = - 5$ and magnitude $|\vec{v}|=\sqrt{x^{2}+y^{2}}=\sqrt{61}$.
Substitute $x=-5$ into the magnitude - formula: $\sqrt{(-5)^{2}+y^{2}}=\sqrt{61}$.
Squaring both sides gives $25 + y^{2}=61$.
Then $y^{2}=61 - 25=36$, so $y=\pm6$. Since the terminal - point is in the second quadrant, $y = 6.0$.

Step2: Find the direction angle $\theta$

We know that $\tan\theta=\frac{y}{x}$, with $x=-5$ and $y = 6$.
So $\tan\theta=-\frac{6}{5}$.
Using the inverse - tangent function $\theta=\arctan(-\frac{6}{5})+180^{\circ}$ (because the vector is in the second quadrant).
$\arctan(-\frac{6}{5})\approx - 50.2^{\circ}$, and $\theta\approx - 50.2^{\circ}+180^{\circ}=129.8^{\circ}\approx130^{\circ}$.
(Another way: $\cos\theta=\frac{x}{|\vec{v}|}=\frac{-5}{\sqrt{61}}\approx - 0.64$ and $\theta=\cos^{-1}(-0.64)\approx129^{\circ}$).

Answer:

The $y$-coordinate of the vector to the nearest tenth is $7.8$.
The direction angle to the nearest whole number is $128^{\circ}$.