Question

Draw a circle of any radius. From a point outside the circle, draw tangents to the circle inclined at 45${}^{\circ }$ to each other.

Answer 1

Steps of construction :

Note: The quadrilateral PMON has the sum of all its angles as ${360}^{\circ }$ but we know 3 angles angle M and angle N are ${90}^{\circ }$ and angle P is ${45}^{\circ }$. So, angle O has to be ${135}^{\circ }$.

1. Construct a circle of any convenient radius.

2. Draw a radius OM and create $\angle$ MON = ${135}^{\circ }$.

3. At M and N, construct two tangents.

4. Let these tangents intersect at P.

5. $\angle$ MPN = ${45}^{\circ }$



So, to draw the given tangent, draw a tangent from any point on the circle. Join it to a centre and draw another radius inclined to the first at an angle of ${135}^{\circ }$ . Draw another tangent from this point on the circle. The point at which these 2 tangents intersect is the point P outside the circle from which the tangents drawn to the circle are inclined at an angle of ${45}^{\circ }$ .