Answer :
To determine the angle that the resultant vector [tex]\( R \)[/tex] makes with the horizontal axis, we can use the given ratio of [tex]\(\frac{R_x}{R_y} = 2\)[/tex].
1. Identify the relationship:
Given [tex]\(\frac{R_x}{R_y} = 2\)[/tex], we can write:
[tex]\[ R_x = 2R_y \][/tex]
2. Use the tangent function:
The angle [tex]\(\theta\)[/tex] that the resultant vector [tex]\(R\)[/tex] makes with the horizontal axis can be found using the tangent function:
[tex]\[ \tan(\theta) = \frac{R_y}{R_x} \][/tex]
3. Substitute the values:
Substitute [tex]\(R_x = 2R_y\)[/tex] into the tangent equation:
[tex]\[ \tan(\theta) = \frac{R_y}{2R_y} = \frac{1}{2} \][/tex]
4. Calculate the angle in radians:
Find the angle [tex]\(\theta\)[/tex] by taking the arctangent of [tex]\(\frac{1}{2}\)[/tex]:
[tex]\[ \theta = \arctan\left(\frac{1}{2}\right) \][/tex]
5. Convert from radians to degrees:
The arctangent of [tex]\(\frac{1}{2}\)[/tex] in radians can be converted to degrees using the conversion factor [tex]\(180^\circ/\pi\)[/tex]:
[tex]\[ \theta \approx 26.57^\circ \][/tex]
Therefore, the angle that the resultant vector [tex]\(R\)[/tex] makes with the horizontal is approximately:
[tex]\[ \boxed{26.56^\circ} \][/tex]
1. Identify the relationship:
Given [tex]\(\frac{R_x}{R_y} = 2\)[/tex], we can write:
[tex]\[ R_x = 2R_y \][/tex]
2. Use the tangent function:
The angle [tex]\(\theta\)[/tex] that the resultant vector [tex]\(R\)[/tex] makes with the horizontal axis can be found using the tangent function:
[tex]\[ \tan(\theta) = \frac{R_y}{R_x} \][/tex]
3. Substitute the values:
Substitute [tex]\(R_x = 2R_y\)[/tex] into the tangent equation:
[tex]\[ \tan(\theta) = \frac{R_y}{2R_y} = \frac{1}{2} \][/tex]
4. Calculate the angle in radians:
Find the angle [tex]\(\theta\)[/tex] by taking the arctangent of [tex]\(\frac{1}{2}\)[/tex]:
[tex]\[ \theta = \arctan\left(\frac{1}{2}\right) \][/tex]
5. Convert from radians to degrees:
The arctangent of [tex]\(\frac{1}{2}\)[/tex] in radians can be converted to degrees using the conversion factor [tex]\(180^\circ/\pi\)[/tex]:
[tex]\[ \theta \approx 26.57^\circ \][/tex]
Therefore, the angle that the resultant vector [tex]\(R\)[/tex] makes with the horizontal is approximately:
[tex]\[ \boxed{26.56^\circ} \][/tex]
