If two tangents inclined at an angle $$60^\circ$$ are drawn to a circle of radius 3 cm, then length of each tangent is equal to
TPO is a right-angled triangle.
So $$\sin\left(30\right)=\frac{OP}{TO}=\frac{3}{TO}\ =>TO=6$$
Applying Pythagoras theorem in the triangle TPO, we get,
$$PT^2=TO^2-OP^2=36-9=27$$
PT= 3$$\sqrt{\ 3}$$ cm
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