$\overline{JT}$ is tangent to $\odot$$O$ at $T$. Complete.

If $OT=6$ and $JO=10$, then $JT=?$

The statement here given is,

If $OT=6$and $JO=10$, then $JT=\_\_\_\_$

The figure:

$JT$is a tangent to circle with centre $O$ at $T$.

According to theorem $9·2$,

Tangent of the circle is perpendicular to the radius of the circle.

According to theorem $8·2$,

The Square of the hypotenuse of right triangle is equal to the sum of the squares of the sides in a right triangle.

It is given that, $JT$is a tangent to circle with centre $O$at $T$.

So, by theorem $9·2$, it can be said that $\overline{JT}$is perpendicular to $\overline{OT}$.

That is,

$m\angle OTJ=90$

Let, $JT=x$

Consider the right triangle$\Delta OTJ$.

Now, $\overline{JO}$ is the hypotenuse and has length 10and the lengths of the legs are $OT=6$ and $JT=x$.

So, by theorem $8·2$,

$\begin{array}{c}J{O}^2=O{T}^2+J{O}^2\\ {10}^2=6^2+x^2\\ 100=36+x^2\end{array}$

$\begin{array}{c}100-36=x^2\\ 64=x^2\end{array}$

$\begin{array}{c}\sqrt{x^2}=\sqrt{64}\\ x=8\end{array}$

So,$JT=8$

Therefore, the complete statement is,

If $OT=6$ and $JO=10$, then $JT=8$.