Question

Question 4
Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of ${60}^{\circ }$.

Answer 1

We know that radius of the circle is perpendicular to the tangents.
Sum of all the 4 angles of quadrilateral $={360}^{\circ }$
$\therefore$ Angle between the radius $\left(\angle O\right)={360}^{\circ }-({90}^{\circ }+{90}^{\circ }+{60}^{\circ })={120}^{\circ }$
Steps of Construction:
Step I:A point Q is taken on the circumference of the circle and OQ is joined. OQ is radius of the circle.
Step II:Draw another radius OR making an angle equal to 120° with the previous one.

Step III:A point P is taken outside the circle. QP and PR are joined which are perpendicular to OQ and OR respectively.
Thus, QP and PR are the required tangents inclined to each other at an angle of ${60}^{\circ }$.
Justification:
Sum of all angles in the quadrilateral PQOR = 360°
$\angle QOR+\angle ORP+\angle OQR+\angle RPQ={360}^{\circ }$
$\Rightarrow {120}^{\circ }+{90}^{\circ }+{90}^{\circ }+\angle RPQ={360}^{\circ }$
$\Rightarrow \angle RPQ={360}^{\circ }-{300}^{\circ }$
$\Rightarrow \angle RPQ={60}^{\circ }$
Hence, QP and PR are tangents inclined to each other at an angle of ${60}^{\circ }$.