# Find the angle subtended at the origin by the line segment

Question:

Find the angle subtended at the origin by the line segment whose end points are (0, 100) and (10, 0).

Solution:

The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula

$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$

In a right angled triangle the angle opposite the hypotenuse subtends an angle of .

Here let the given points be A(0,100), B(10,0). Let the origin be denoted by O(0,0).

Let us find the distance between all the pairs of points

$A B=\sqrt{(0-10)^{2}+(100-0)^{2}}$

$=\sqrt{(-10)^{2}+(100)^{2}}$

$=\sqrt{100+10000}$

$A B=\sqrt{10100}$

\begin{aligned} A O &=\sqrt{(0-0)^{2}+(100-0)^{2}} \\ &=\sqrt{(0)^{2}+(100)^{2}} \end{aligned}

$=\sqrt{(0)^{2}+(100)^{2}}$

$A O=\sqrt{10000}$

$B O=\sqrt{(10-0)^{2}+(0-0)^{2}}$

$=\sqrt{(10)^{2}+(0)^{2}}$

$B O=\sqrt{100}$

Here we can see that $A B^{2}=A O^{2}+B O^{2}$.

So, is a right angled triangle with ‘AB’ being the hypotenuse. So the angle opposite it has to be . This angle is nothing but the angle subtended by the line segment ‘AB’ at the origin.

Hence the angle subtended at the origin by the given line segment is .