Chapter 12.2, Problem 28WE

To determine

Find the length of x .

Answer to Problem 28WE

  $x=24-8\sqrt{3}$

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, $x=BC-CD$

In $\Delta ABC$ ,

  $\angle A=45°$

AC = length of side adjacent to $\angle A$ =24 units

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

AB = length of the hypotenuse

  $\begin{array}{l}\tan \angle A=\frac{\text{length of side opposite to }\angle A}{\text{length of side adjacent to }\angle A}=\frac{BC}{AC}\\ \text{⇒}AC\tan \angle A=BC\\ \text{⇒}BC=24\tan 45°\mathrm{.............}[\angle A=45°,AC=24]\\ \text{⇒}BC=\mathrm{24...........}(1)\mathrm{..............}[\tan 45°=1]\end{array}$

In $\Delta ADC$ ,

  $\angle DAC=30°$

AC = length of side adjacent to $\angle A$ =24 units

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

AD = length of the hypotenuse

  $\begin{array}{l}\tan \angle A=\frac{\text{length of side opposite to }\angle A}{\text{length of side adjacent to }\angle A}=\frac{DC}{AC}\\ \text{⇒}AC\tan \angle A=DC\\ \text{⇒}DC=24\tan 30°\mathrm{.............}[\angle A=30°,AC=24]\\ \text{⇒}DC=24\frac{\sqrt{3}}{3}\mathrm{..................}[\tan 30°=\frac{\sqrt{3}}{3}]\\ \text{⇒}DC=8\sqrt{3}\end{array}$

From equation (1) and (2):

  $BC=24$

  $DC=8\sqrt{3}$

  $\begin{array}{l}x=BC-DC\\ \text{⇒}x=24-8\sqrt{3}\end{array}$