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Question Number 169038 by mr W last updated on 23/Apr/22

Commented by mr W last updated on 23/Apr/22

$$oldquestionreposted$$

Answered by mr W last updated on 23/Apr/22

Commented by cortano1 last updated on 23/Apr/22

$$BCisdiameter?$$

Commented by mr W last updated on 23/Apr/22

$$BCcanbe,but{mustn}^{\prime }tbediameter.$$

Commented by mr W last updated on 23/Apr/22

$$\frac{PS}{AP}=\frac{PB}{CP}$$$$\Rightarrow PS=\frac{2\times (5+4)}{x}=\frac{18}{x}$$$$\frac{QT}{QB}=\frac{AQ}{CQ}$$$$\Rightarrow QT=\frac{4\times (2+5)}{y}=\frac{28}{y}$$$$\cos \mathrm{\angle }APS=\frac{2^2+{\left(\frac{18}{x}\right)}^2-5^2}{2\times 2\times \frac{18}{x}}=\frac{5^2+x^2-y^2}{2\times 5\times x}$$$$\Rightarrow 47x^2-12y^2=240...\left(i\right)$$$$\cos \mathrm{\angle }TQB=\frac{4^2+{\left(\frac{28}{y}\right)}^2-2^2}{2\times 4\times \frac{28}{y}}=\frac{5^2+y^2-x^2}{2\times 5\times y}$$$$\Rightarrow 13y^2-28x^2=280...\left(ii\right)$$$$from\left(i\right)and\left(ii\right):$$$$x^2=\frac{1296}{55}$$$$y^2=\frac{3976}{55}$$$$letST=s$$$$\cos \mathrm{\angle }PCQ=\frac{x^2+y^2-5^2}{2xy}=\frac{{(x+\frac{18}{x})}^2+{(y+\frac{28}{y})}^2-s^2}{2(x+\frac{18}{x})(y+\frac{28}{y})}$$$$\Rightarrow s^2=x^2{(1+\frac{18}{x^2})}^2+y^2{(1+\frac{28}{y^2})}^2-(x^2+y^2-5^2)(1+\frac{18}{x^2})(1+\frac{28}{y^2})$$$$\Rightarrow s^2=\frac{1296}{55}{(1+\frac{18\times 55}{1296})}^2+\frac{3976}{55}{(1+\frac{28\times 55}{3976})}^2-(\frac{1296}{55}+\frac{3976}{55}-5^2)(1+\frac{18\times 55}{1296})(1+\frac{28\times 55}{3976})$$$$=\frac{625}{16}$$$$\Rightarrow s=\frac{25}{4}=6.25$$

Commented by Tawa11 last updated on 23/Apr/22

$$\mathrm{Great}\mathrm{sir}.$$

Answered by mr W last updated on 24/Apr/22

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