Question

Draw a circle of radius of $5{\rm{ cm}}$. Draw two tangents to the circle so that angle between tangents is $45^\circ $.

Answer 1

Hint: This question is based on Geometry construction and we have to construct a circle of given radius and then draw two tangents to the circle in such a way that the angle between both the tangents is $45^\circ $. Also, we know that a tangent always makes an angle of $90^\circ $ between the tangent line and the line joining the tangent contact point and the centre of the circle. So, we use the geometric property of a quadrilateral to construct the tangents with the angle of $45^\circ $ between them.

Complete step-by-step answer:
Given:
The radius of the circle $OB = {\rm{5 cm}}$.
And, the angle between two tangents to the circle $ = 45^\circ $

The step by step construction is given below:
i) First take a radius of ${\rm{5 cm}}$ using the ruler scale and draw a circle taking O as the centre of the circle. So, $OB = {\rm{5 cm}}$
ii) Taking O as the centre draw an angle $\angle AOB$ such that $\angle AOB = 135^\circ $
iii) Now at point A draw a line making an angle of $90^\circ $, similarly at the point B draw another line making an angle of $90^\circ $.
iv) The point at which these two lines meet each other, name it as point P, this point P is outside the circle.
Therefore, AP and BP are the required two tangents on a circle of radius $5{\rm{ cm}}$. making an angle of $45^\circ $ between each other.

Note: It should be noted that AOBP is a Quadrilateral and from the geometric property of a quadrilateral “the sum of the interior angles of a quadrilateral is always equals to $360^\circ $”.
So, the interior angles of AOBP are-
$\Rightarrow \angle AOB = 135^\circ $
$\Rightarrow \angle OAP = 90^\circ $
$\Rightarrow \angle OBP = 90^\circ $
And,
$\Rightarrow \angle APB = 45^\circ $
So, the sum of the interior angles is given by-
$
\Rightarrow \angle AOB + \angle OAP + \angle OBP + \angle APB = 135^\circ + 90^\circ + 90^\circ + 45^\circ \\
 = 360^\circ
$