Chapter 11.2, Problem 28PSC

To determine

To find the area of triangle ACB.

Answer to Problem 28PSC

The area of triangle ACBis $10.14$ .

Explanation of Solution

Given:

In triangle ADB,

  $AC=8$

Formula Used:

  $\begin{array}{l}Areaoftriangle=\frac{1}{2}\ast b\ast h\\ b=Base,\\ h=Height\end{array}$

Calculation:

  

  $\begin{array}{l}\angle CAB={180}^o-({120}^o+{45}^o)\\ \angle CAB={180}^o-({165}^o)\\ \angle CAB={15}^o\end{array}$

Extend $\overline{CB}$ and draw a perpendicular from point A to from right angle $\Delta ABD$

  $\begin{array}{l}\angle DCA={180}^o-{120}^o={60}^o\\ \Delta CDAisa30^{o}60^{o}90^o\Delta .\end{array}$

  $\begin{array}{l}DC=\frac{1}{2}(AC)\\ DC=\frac{1}{2}(8)=4\\ DA=(DC)\sqrt{3}\\ =4\sqrt{3}\\ \angle DAC+\angle CAB=\angle DAB\\ {30}^{\circ }+{15}^{\circ }=\angle DAB\\ \angle DAB={45}^{\circ }\end{array}$

  $\Delta DAB\text{ is a}{\text{45}}^{\text{o}}{\text{45}}^{\text{o}}{\text{90}}^{\text{o}}\Delta .$

If corresponding angles are congruent then there sides are also congruent.

  $\begin{array}{l}\overline{DA}\cong \overline{DB}\\ DB=4\sqrt{3}\\ CB=DB-DC\\ CB=4\sqrt{3}-4\end{array}$

  $\begin{array}{l}A(\Delta ACB)=\frac{1}{2}*AD*CB\\ =\frac{1}{2}*4\sqrt{3}\ast (4\sqrt{3}-4)\\ =\frac{1}{2}(48-16\sqrt{3})\\ =24-8\sqrt{3}\\ =10.14\end{array}$