
A vector of magnitude 10 has its rectangular components as 8 and 6 along x and y axes. Find the angle it makes with these axes.
Answer
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Hint:To solve this type of question we will use the concept of the vector component. In physics, Vector is a quantity that has both direction and magnitude. It is represented by an arrow. In which the direction is the same as that of the quantity and length is proportional to the magnitude of quantity. Using trigonometric equations like sin, cos, tan, etc. we can solve this type of question.
Formula used:$\tan \theta = \dfrac{{opposite}}{{adjacent}}$sides
Complete step-by-step answer:
Let us write first the information given in the question.
Let A be some vector then,\[\left| {\vec A} \right| = 10\] ${A_x} = 8$ and ${A_y} = 6$
We have to find the angle it makes with x and y axes.
Let us consider that it makes an angle α with x-axes and β with y-axes, as shown in the figure.
Now using the diagram let us find them $\tan \alpha $.
From the trigonometry we know that,
$\tan \alpha $= $\dfrac{{{A_y}}}{{{A_x}}}$
Where, ${A_y}$=Opposite side to the angle
\[{A_x}\]=Adjacent side to the angle
Let us substitute the values for ${A_y}$ and \[{A_x}\]in the $\tan \alpha $and we get the following,
$\tan \alpha $= $\dfrac{6}{8} = \dfrac{3}{4}$
Let us calculate the value for$\alpha $,
$\tan \alpha = \dfrac{3}{4} \Rightarrow \alpha = {\tan ^{ - 1}}\dfrac{3}{4}$
On solving for the $\alpha $we get,
$\alpha $= \[{36.869^0}\]
Angle made by vector with x-axis = $\alpha $= \[{36.869^0}\] \[\]
Similarly, let us now calculate tanβ using the figure.
$\tan \beta = \dfrac{{{A_x}}}{{{A_y}}}$
Let us calculate it for the value of β.
$\tan \beta = \dfrac{4}{3} \Rightarrow \beta = {\tan ^{ - 1}}\dfrac{4}{3}$
Now let us write the angle of tan for which it gives 4/3.
β=53.130 degree.
Hence, required angles are 36.8690 and 53.1300.
Note:
There is another method to solve it.
First step to calculate either α or β by making use of the figure.
After that to find the other angle we can subtract it from 90 degrees.
i.e.${90^0} - {36.869^0}$=${53.130^0}$
Vector is a physical quantity which has both direction and magnitude.
Formula used:$\tan \theta = \dfrac{{opposite}}{{adjacent}}$sides
Complete step-by-step answer:
Let us write first the information given in the question.
Let A be some vector then,\[\left| {\vec A} \right| = 10\] ${A_x} = 8$ and ${A_y} = 6$
We have to find the angle it makes with x and y axes.
Let us consider that it makes an angle α with x-axes and β with y-axes, as shown in the figure.
Now using the diagram let us find them $\tan \alpha $.
From the trigonometry we know that,
$\tan \alpha $= $\dfrac{{{A_y}}}{{{A_x}}}$
Where, ${A_y}$=Opposite side to the angle
\[{A_x}\]=Adjacent side to the angle
Let us substitute the values for ${A_y}$ and \[{A_x}\]in the $\tan \alpha $and we get the following,
$\tan \alpha $= $\dfrac{6}{8} = \dfrac{3}{4}$
Let us calculate the value for$\alpha $,
$\tan \alpha = \dfrac{3}{4} \Rightarrow \alpha = {\tan ^{ - 1}}\dfrac{3}{4}$
On solving for the $\alpha $we get,
$\alpha $= \[{36.869^0}\]
Angle made by vector with x-axis = $\alpha $= \[{36.869^0}\] \[\]
Similarly, let us now calculate tanβ using the figure.
$\tan \beta = \dfrac{{{A_x}}}{{{A_y}}}$
Let us calculate it for the value of β.
$\tan \beta = \dfrac{4}{3} \Rightarrow \beta = {\tan ^{ - 1}}\dfrac{4}{3}$
Now let us write the angle of tan for which it gives 4/3.
β=53.130 degree.
Hence, required angles are 36.8690 and 53.1300.
Note:
There is another method to solve it.
First step to calculate either α or β by making use of the figure.
After that to find the other angle we can subtract it from 90 degrees.
i.e.${90^0} - {36.869^0}$=${53.130^0}$
Vector is a physical quantity which has both direction and magnitude.
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