In a triangle ABC with usual notations,if ang | vedclass

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(C) Given the expression $E = \left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right)$.Simplifying the terms,we get $E = \left(\f

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$\sqrt{3}+2$

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(C) Given the expression $E = \left(1+\frac{a}{c}+\frac{b}{c}ight)\left(1+\frac{c}{b}-\frac{a}{b}ight)$.Simplifying the terms,we get $E = \left(\frac{c+a+b}{c}ight)\left(\frac{b+c-a}{b}ight)$.This can be written as $E = \frac{(b+c)+a}{c} 	imes \frac{(b+c)-a}{b} = \frac{(b+c)^2 - a^2}{bc}$.Expanding the numerator,we get $E = \frac{b^2+c^2+2bc-a^2}{bc} = \frac{b^2+c^2-a^2}{bc} + 2$.Using the Law of Cosines,$\cos A = \frac{b^2+c^2-a^2}{2bc}$,so $b^2+c^2-a^2 = 2bc \cos A$.Substituting this into the expression,$E = \frac{2bc \cos A}{bc} + 2 = 2 \cos A + 2$.Given $\angle A = 30^{\circ}$,we have $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$.Thus,$E = 2 \left(\frac{\sqrt{3}}{2}ight) + 2 = \sqrt{3} + 2$.

In a triangle $ABC$ with usual notations,if $\angle A = 30^{\circ}$,then the value of $\left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right)=$

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