Chapter 12.2, Problem 26WE

To determine

Find the length of x .

Answer to Problem 26WE

  $x=\frac{5}{2}\cdot \left(3\sqrt{2}+\sqrt{6}\right)$

Explanation of Solution

Given:

  

Formula Used:

  $\begin{array}{l}\sin \theta =\frac{\text{length of side opposite to }\theta }{\text{length of the hypotenuse}}=\frac{a}{c}\\ \text{⇒}c\sin \theta =a\end{array}$

  $\begin{array}{l}\cos \theta =\frac{\text{length of side adjacent to }\theta }{\text{length of the hypotenuse}}=\frac{b}{c}\\ \text{⇒}c\cos \theta =b\end{array}$

  $\tan \theta =\frac{\text{length of side opposite to }\theta }{\text{length of side adjacent to }\theta }$

Calculation:

Split the triagle in two right-angled triangles.

  

  

So, $x=AC+CD$

In $\Delta ABC$ ,

  $\angle A=45°$

AC = length of side adjacent to $\angle A$

BC = length of side opposite to $\angle A$

AB = length of the hypotenuse = 15 units

  $\begin{array}{l}AB\cos \angle A=AC\\ \text{⇒}15\cos 45°=AC\mathrm{.........}[\angle A=45°,AB=15]\\ \text{⇒}15\cdot \frac{\sqrt{2}}{2}=AC\mathrm{..............}[\cos 45°=\frac{\sqrt{2}}{2}]\\ \text{⇒}AC=\frac{15\sqrt{2}}{2}\mathrm{.......................}(1)\end{array}$

Find BC by using Pythagorean Theorem:

  $\begin{array}{l}B{C}^2+A{C}^2=A{B}^2\\ \text{⇒}BC=\sqrt{A{B}^2-A{C}^2}=\sqrt{225-\frac{225}{2}}=\frac{15\sqrt{2}}{2}\\ \text{⇒}BC=\frac{15\sqrt{2}}{2}\end{array}$

In $\Delta BCD$ ,

  $BC=\frac{15\sqrt{2}}{2},\angle D=60°$

  $\begin{array}{l}\tan \angle D=\frac{\text{length of side opposite to }\angle \text{D}}{\text{length of side adjacent to }\angle \text{D}}=\frac{BC}{CD}\\ \text{⇒}CD\tan \angle D=BC\\ \text{⇒}CD=\frac{BC}{\tan 60°}\mathrm{.............}[\angle D=60°]\\ \text{⇒}CD=\frac{\frac{15\sqrt{2}}{2}}{\sqrt{3}}\mathrm{..............}[\tan 60°=\sqrt{3}]\\ \text{⇒}CD=\frac{15\sqrt{2}}{2\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}\mathrm{.......}[rationalize]\\ \text{⇒}CD=\frac{15\sqrt{6}}{6}=\frac{5\sqrt{6}}{2}\mathrm{.................}(2)\end{array}$

From equation (1) and (2):

  $AC=\frac{15\sqrt{2}}{2}$

  $CD=\frac{5\sqrt{6}}{2}$

  $\begin{array}{l}x=AC+CD\\ \text{⇒}x=\frac{15\sqrt{2}}{2}+\frac{5\sqrt{6}}{2}\\ \text{⇒}x=\frac{5}{2}\cdot \left(3\sqrt{2}+\sqrt{6}\right)\end{array}$