Question

QN1 Angle
between à (25m) i +(45m) i
Positive x-axis is:
and
the
(A) 299
(B) 61°
(c) 151°
(0) 2090​

Answer 1

Answer:

A vector is defined as a quantity having both, the magnitude and the direction. In the Cartesian coordinate system, a vector is written in the form of the unit

(

^

i

,

^

j

)

vectors. The vector

→

A

can be written as:

→

A

=

A

x

^

i

+

A

y

^

j

Here

A

x

^

i

is the horizontal component and

A

y

^

j

is the vertical component.

The magnitude of the vector

→

A

can be determined as:

A

=

√

(

A

x

)

2

+

(

A

y

)

2

Also, the direction of the vector can be calculated as,

θ

=

tan

−

1

A

y

A

x

Answer and Explanation:

Given:

→

A

=

(

−

25

m

)

^

i

+

(

45

m

)

^

j

Determining the angle between the given vector and the positive x-axis:

θ

=

180

−

tan

−

1

(

45

25

)

=

180

−

60

∘

56

′

43

′′

=

119

∘

3

′

16

′′

Therefore, the angle between the given vector and the positive x-axis "119 degrees 3 minutes 16 seconds".

Answer 2

Answer:

The angle between the given vector $\vec A=25\hat i+45\hat j$ and the positive x-axis is $61^{0}$

Explanation:

The proper question is-"What is the angle between the vector $\vec A=25\hat i+45\hat j$ and the positive x-axis?

The given vector is $\vec A=25\hat i+45\hat j$ and lets assume that, the angle made by the vector with positive x-axis be $\theta$.

We know that If a vector $\vec P=x\hat i+y\hat j$ makes an angle $\theta$ with x-axis then its components are as follows

$x=Pcos\theta\\y=Psin\theta$

Therefore, for the given question we can write the components based on the above relations.

$25=Acos\theta\\= > cos\theta=\frac{25}{A}=\frac{25}{\sqrt{25^2+45^2} } =0.485\\= > \theta=cos^{-1}(0.485)=60.9^{0}\\\approx 61^{0}$

Therefore,

The angle between the given vector $\vec A=25\hat i+45\hat j$ and the positive x-axis is found to be $61^{0}$

#SPJ3