To express the vector [tex]\( \mathbf{t} = -5i + 12j \)[/tex] in trigonometric form, we need to find its magnitude and the angle it forms with the positive x-axis (measured counterclockwise).
### Step-by-Step Solution:
1. Calculate the Magnitude:
The magnitude (or length) of the vector [tex]\(\mathbf{t}\)[/tex] can be found using the Pythagorean theorem:
[tex]\[
|\mathbf{t}| = \sqrt{(-5)^2 + 12^2}
\][/tex]
Plugging in the values,
[tex]\[
|\mathbf{t}| = \sqrt{25 + 144} = \sqrt{169} = 13
\][/tex]
2. Determine the Angle:
The angle [tex]\(\theta\)[/tex] that the vector makes with the positive x-axis can be found using the arctangent function. Specifically, we use the arctangent of the ratio of the y-component to the x-component of the vector:
[tex]\[
\theta = \arctan\left(\frac{12}{-5}\right)
\][/tex]
Since we are dealing with a vector in the 2nd quadrant (where [tex]\(x < 0\)[/tex] and [tex]\(y > 0\)[/tex]), the angle [tex]\(\theta\)[/tex] is:
[tex]\[
\theta = \arctan\left(\frac{12}{-5}\right) \approx 1.965587446494658 \text{ radians}
\][/tex]
To convert this angle to degrees:
[tex]\[
\theta_{\text{degrees}} = \theta \times \left(\frac{180}{\pi}\right) \approx 112.61986494804043^\circ
\][/tex]
3. Express the Vector in Trigonometric Form:
A vector in trigonometric form can be written as:
[tex]\[
\mathbf{t} = |\mathbf{t}| (\cos \theta \mathbf{i} + \sin \theta \mathbf{j})
\][/tex]
Substituting the magnitude and angle we found:
[tex]\[
\mathbf{t} = 13 (\cos 112.61986494804043^\circ \mathbf{i} + \sin 112.61986494804043^\circ \mathbf{j})
\][/tex]
In summary, the trigonometric form of the vector [tex]\(\mathbf{t} = -5i + 12j\)[/tex] is:
[tex]\[
\mathbf{t} = 13 (\cos 112.61986494804043^\circ \mathbf{i} + \sin 112.61986494804043^\circ \mathbf{j})
\][/tex]
This form concisely captures the magnitude and direction of the vector.
