Chapter 9.2, Problem 2MRE

Solution

To find: $AB$

  $AB=9\sqrt{2}$

Given:

The figure:

Concept Used:

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

Explanation:

Consider the given figure:

From the above figure it is clear that this triangle is a $45°-45°-90°$ triangle. So,

  $m\angle CAB=m\angle CBA=45$

Consider the right triangle $△OTJ$ .

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

Now perform cross multiplication,

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