Pythagoras Theorem and Cosines Law

ABCD is a square . If $\displaystyle BC=1\ \ \text{and  }a+b+c=2$ , find x .

Solution

By the Pythagoras theorem ,

$\displaystyle {{a}^{2}}+{{b}^{2}}={{c}^{2}}$

$\displaystyle a+b+c=2$

$\displaystyle {{(a+b)}^{2}}={{(2-c)}^{2}}$

$\displaystyle {{a}^{2}}+2ab+{{b}^{2}}=4-4c+{{c}^{2}}$

$\displaystyle ab=2-2c\ \ \ (\because {{a}^{2}}+{{b}^{2}}={{c}^{2}})$

By the law of cosines ,

$\displaystyle \cos x=\frac{{2-2a+{{a}^{2}}+2-2b+{{b}^{2}}-{{c}^{2}}}}{{2\sqrt{{(2-2a+{{a}^{2}})(2-2b+{{b}^{2}})}}}}$

$\displaystyle \ \ \ \ \ \ \ \ =\frac{{2-a-b}}{{\sqrt{{4-4b+2{{b}^{2}}-4a+4ab-2a{{b}^{2}}+2{{a}^{2}}-2{{a}^{2}}b+{{a}^{2}}{{b}^{2}}}}}}$

$\displaystyle \ \ \ \ \ \ \ \ =\frac{c}{{\sqrt{{4(1-b-a+ab)+2({{a}^{2}}+{{b}^{2}})-2ab(a+b)+{{{(2-2c)}}^{2}}}}}}$

$\displaystyle \ \ \ \ \ \ \ \ =\frac{c}{{\sqrt{{4(1-c)+2{{c}^{2}}-2(4-6c+2{{c}^{2}})+4-8c+4{{c}^{2}}}}}}$

$\displaystyle \ \ \ \ \ \ \ \ =\frac{c}{{\sqrt{{8-12c+6{{c}^{2}}-8+12c-4{{c}^{2}}}}}}$

$\displaystyle \ \ \ \ \ \ \ \ =\frac{c}{{\sqrt{{2{{c}^{2}}}}}}$

$\displaystyle \ \ \ \ \ \ \ \ =\frac{1}{{\sqrt{2}}}=\cos {{45}^{\circ }}\ $

$\displaystyle \therefore \ \ x={{45}^{\circ }}$