Question

in the figure below, the segments \\( \overline { t u } \\) and \\( \overline { t v } \\) are tangent to the circle centered at \\( o \\). given that \\( t v = 2.4 \\) and \\( o t = 3 \\), find \\( o u \\).

Explanation:

Step1: Use the tangent - radius property

Since \(TU\) and \(TV\) are tangents to the circle centered at \(O\), \(OU\perp TU\) (a tangent to a circle is perpendicular to the radius at the point of tangency). So, \(\triangle{OTU}\) is a right - triangle with \(\angle{OUT} = 90^{\circ}\).

Step2: Apply the Pythagorean theorem

In right - triangle \(\triangle{OTU}\), by the Pythagorean theorem \(OT^{2}=OU^{2}+TU^{2}\). We know that \(OT = 3 + OU\) (assuming \(OT\) is composed of the radius \(OU\) and another segment of length \(3\), and \(TU = 24\)). Let \(OU=x\), then \(OT=x + 3\).
Substituting into the Pythagorean formula: \((x + 3)^{2}=x^{2}+24^{2}\).
Expand \((x + 3)^{2}\): \(x^{2}+6x+9=x^{2}+576\).
Subtract \(x^{2}\) from both sides: \(6x+9 = 576\).
Subtract \(9\) from both sides: \(6x=576 - 9=567\).
Divide both sides by \(6\): \(x=\frac{567}{6}=94.5\).

Answer:

\(94.5\)