Chapter 12.2, Problem 27WE

To determine

Find the length of x .

Answer to Problem 27WE

  $x=6(1+\sqrt{3})$

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 triangle in two right-angled triangles.

  

  

So, $x=AC+CD$

In $\Delta ABC$ ,

  $\angle A=30°$

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

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

AB = length of the hypotenuse = 12 units

  $\begin{array}{l}AB\cos \angle A=AC\\ \text{⇒}12\cos 30°=AC\mathrm{.........}[\angle A=30°,AB=12]\\ \text{⇒}12\cdot \frac{\sqrt{3}}{2}=AC\mathrm{..............}[\cos 30°=\frac{\sqrt{3}}{2}]\\ \text{⇒}AC=6\sqrt{3}\mathrm{.......................}(1)\end{array}$

Find BC by using Pythagorean Theorem:

  $\begin{array}{l}B{C}^2+A{C}^2=A{B}^2\\ \text{⇒}BC=\sqrt{A{B}^2-A{C}^2}=\sqrt{144-108}=6\\ \text{⇒}BC=6\end{array}$

In $\Delta BCD$ ,

  $BC=6,\angle D=45°$

  $\begin{array}{l}\tan \angle D=\frac{\text{length of side opposite to }\angle \text{D}}{\text{length of side adjacent to }\angle \text{D}}=\frac{BC}{CD}\\ \text{⇒}CD\tan \angle D=BC\\ \text{⇒}CD=\frac{6}{\tan 45°}\mathrm{.............}[\angle D=45°]\\ \text{⇒}CD=\frac{6}{1}\mathrm{..............}[\tan 45°=1]\\ \text{⇒}CD=\mathrm{6.................}(2)\end{array}$

From equation (1) and (2):

  $AC=6\sqrt{3}$

  $CD=6$

  $\begin{array}{l}x=AC+CD\\ \text{⇒}x=6\sqrt{3}+6\\ \text{⇒}x=6(1+\sqrt{3})\end{array}$