Questions: What is the angle between the given vector and the positive direction of the x-axis? (Round your answer to the nearest degree.) i+√15 j

Find the angle between the vector \( i + \sqrt{15} j \) and the positive direction of the \(x\)-axis.

Identify the components of the vector.

The vector is \( i + \sqrt{15} j \), so the \(x\)-component is \( 1 \) and the \(y\)-component is \( \sqrt{15} \).

Use the arctangent function to find the angle.

The angle \( \theta \) is given by:
\[ \theta = \arctan\left(\frac{y\text{-component}}{x\text{-component}}\right) = \arctan\left(\frac{\sqrt{15}}{1}\right) = \arctan(\sqrt{15}). \]

Calculate the value of \( \arctan(\sqrt{15}) \).

Using a calculator, \( \arctan(\sqrt{15}) \approx 75.52^\circ \).

Round the angle to the nearest degree.

Rounding \( 75.52^\circ \) to the nearest degree gives \( 76^\circ \).

The angle between the vector and the positive direction of the \(x\)-axis is \( \boxed{76^\circ} \).

The angle between the vector and the positive direction of the \(x\)-axis is \( \boxed{76^\circ} \).