Circle: Area of triangle and cosines law

Given : In circle O , $\displaystyle AB=\sqrt{{41}}\ ,\ AC=5\ ,\ BC=8$ and $\displaystyle A{{A}^{'}}$ is a diameter and as shown in figure .
Find : area of triangle $\displaystyle {{A}^{'}}BC$ .

Solution

$\displaystyle AB=\sqrt{{41}}\ ,\ AC=5\ ,\ BC=8$

$\displaystyle \text{Draw}\ {{A}^{'}}D\bot BC.$ 

$\displaystyle A{{B}^{2}}+{{A}^{'}}{{B}^{2}}=A{{C}^{2}}+{{A}^{'}}{{C}^{2}}$

$\displaystyle 41+{{A}^{'}}{{D}^{2}}+B{{D}^{2}}=25+{{A}^{'}}{{D}^{2}}+D{{C}^{2}}$

$\displaystyle D{{C}^{2}}-B{{D}^{2}}=16$ 

$\displaystyle 8\ (DC-BD)=16$ 

$\displaystyle \ \ \ \ \ DC-BD=2$

$\displaystyle \therefore \ BD=3\ \ ,\ \ DC=5$ 

$\displaystyle \text{By}\ \text{the}\ \text{law}\ \text{of}\ \text{cosines}\ ,$ 

$\displaystyle \cos \theta =\frac{{A{{B}^{2}}+B{{C}^{2}}-A{{C}^{2}}}}{{2AB.BC}}$

$\displaystyle \ \ \ \ \ \ \ =\frac{{41+64-25}}{{16\sqrt{{41}}}}$ 

$\displaystyle \ \ \ \ \ \ \ =\frac{5}{{\sqrt{{41}}}}$ 

$\displaystyle \sin ({{90}^{\circ }}-\theta )=\frac{5}{{\sqrt{{41}}}}\ \ \ \ \ \left[ {\because \sin ({{{90}}^{\circ }}-\theta )=\cos \theta } \right]$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \frac{{{{A}^{'}}D}}{{{{A}^{'}}B}}=\frac{5}{{\sqrt{{41}}}}\ $

$\displaystyle \ \ \ \ \ \ \ \ \ \ \frac{{{{A}^{'}}{{D}^{2}}}}{{{{A}^{'}}{{B}^{2}}}}=\frac{{25}}{{41}}$

$\displaystyle \ \ \ \ \ \frac{{{{A}^{'}}{{D}^{2}}}}{{{{A}^{'}}{{D}^{2}}+9}}=\frac{{25}}{{41}}$ 

$\displaystyle 41\ {{A}^{'}}{{D}^{2}}=25\ {{A}^{'}}{{D}^{2}}+\ 225$ 

$\displaystyle \ \ \ {{A}^{'}}{{D}^{2}}=\frac{{225}}{{16}}$

$\displaystyle \ \ \ \ {{A}^{'}}D=\frac{{15}}{4}$ 

$\displaystyle \therefore \ \alpha \ (\ \Delta \ {{A}^{'}}BC\ )=\frac{1}{2}\ \times \ BC\ \times \ {{A}^{'}}D$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}\ \times \ 8\ \times \ \frac{{15}}{4}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =15\ \ \text{sq}\ \text{units}$