Question

Given vector A=2i^+3j^, the angle between A and y-axis is

Answer 1

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Answer 2

The angle between A and y-axis is tan⁻¹ (2/3)

Explanation:

$\overrightarrow{A} = 2 \hat{i} + 3 \hat{j}$

Now,

A = $|\overrightarrow{A}|$ = √(2² + 3²) = √(4 + 9) = √(13)

Now, the cosine is given as:

cos β = (Adjacent side)/(Hypotenuse)

On substituting the values, we get,

cos β = 3/√(13)

Now, the sine is given as:

sin β = (Opposite side)/(Hypotenuse)

On substituting the values, we get,

sin β =  2/√(13)

Now, the angle between A and y axis is given as:

tan β = (Opposite side)/(Adjacent side)

On substituting the values, we get,

tan β = 2/3

∴ β = tan⁻¹ (2/3)