If two tangents to a circle of radius 3 cm are inclined to each other…

Let AP and CP be the tangents drawn to a circle

According to the question,

\(\angle\)CPA = \({60}^\circ\)

So, AP = CP

In triangle PAO and PCO,

= AP = CP

= AP = CO = radius of circle

= PO = PO = common side

Triangle PAO and triangle PCO are similar, by SSS criterion,

At the point of contact the radius of a circle is perpendicular to the tangent.

So, \(\angle\)PAO = \(\angle\)PCO = \({90}^\circ\)

Since AO = CO = radius

= \(\angle\)PAO = \(\angle\)PCO

= \(\angle\)CPA = \(\angle\)PAO + \(\angle\)PCO

= 2\(\angle\)PAO = \({60}^\circ\)

= \(\angle\)PAO = \({60}^\circ\)/2

= \(\angle\)PAO = \({30}^\circ\)

In triangle PAO

PAO is aright angle triangle with A at right triangle.

= tan \({30}^\circ\) = \(\frac{AO}{PA}\)

= \(\frac{1}{√3}\) = \(\frac{3}{PA}\)

= AP = 3√3 cm.

Therefore, length of each tangent is 3√3 cm.