Question

Find $AB.$In Exercise $3,$$\overline{CB}$ is tangent to $\odot A.$

Short answer

The length of,$AB=9\sqrt{2}$.

Step-by-step solution

Step 1. Given information.

The figure here given is,

Step 2. Concept Used.

$\cos A=\frac{\text{leg adjacent to}\angle A}{\text{hypotenuse}}$

Step 3. Consider the given figure for further solution.

From the above figure it is clear that this triangle is a $45{}^{\circ }-45{}^{\circ }-90°$triangle.

So,

$m\angle CAB=m\angle CBA=45$

Consider the right triangle $\Delta OTJ$.

$\begin{array}{c}\cos \left(CAB\right)=\frac{\text{legadjacentto}\angle TOJ}{\text{hypotenuse}}\\ \cos 45°=\frac{CA}{AB}\\ \frac{1}{\sqrt{2}}=\frac{9}{AB}\end{array}$

Step 4. Now perform cross multiplication.

$\begin{array}{c}AB=9\times \sqrt{2}\\ AB=9\sqrt{2}\end{array}$

Therefore,the length of,$AB=9\sqrt{2}$.