Chapter 15, Problem 22CR

To determine

To Find: The value of required angle in the given figure.

Answer to Problem 22CR

  $x=3\sqrt{5}$

Explanation of Solution

Given:

  

  $\overline{BD}\text{Angle-bisector of }\angle \text{ABC}\text{.}$

Formula Used:

  $\text{Pythagoras Theorem}:h^2=b^2+p^2$

Angle bisector theorem.

Calculation:

  $\begin{array}{l}\overline{BD}\text{Angle-bisector of }\angle \text{ABC}\text{.}\\ \frac{BC}{AB}=\frac{CD}{AD}\\ A{C}^2+A{B}^2=B{C}^2\\ A{C}^2=B{C}^2-A{B}^2\\ \text{⇒}A{C}^2={10}^2-6^2=100-36=64\\ \text{⇒}AC=\sqrt{64}=8\\ \frac{10}{6}=\frac{8-AD}{AD}\\ \text{⇒}10AD=6\left(8-AD\right)\\ \text{⇒}10AD=48-6AD\\ \text{⇒}10AD+6AD=48\\ \text{⇒}16AD=48\\ \text{⇒}AD=\frac{48}{16}\\ \text{⇒}AD=3\\ B{D}^2=A{D}^2+A{B}^2\\ \text{⇒}{x}^2=3^2+6^2\\ \text{⇒}{x}^2=9+36\\ \text{⇒}x=\sqrt{45}\\ \text{⇒}x=3\sqrt{5}\end{array}$

Conclusion:

Thus the required value $x=3\sqrt{5}$