With usual notation,in triangle ABC,m angle A | vedclass

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(B) Given the expression $E = \left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right)$.This can be rewritten as $E = \left(\frac{

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$2+\sqrt{3}$

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(B) Given the expression $E = \left(1+\frac{a}{c}+\frac{b}{c}ight)\left(1+\frac{c}{b}-\frac{a}{b}ight)$.This can be rewritten as $E = \left(\frac{c+a+b}{c}ight)\left(\frac{b+c-a}{b}ight)$.$E = \frac{(b+c)+a}{c} 	imes \frac{(b+c)-a}{b} = \frac{(b+c)^2 - a^2}{bc}$.Expanding the numerator: $E = \frac{b^2 + c^2 + 2bc - a^2}{bc} = \frac{b^2 + c^2 - a^2}{bc} + 2$.Using the Law of Cosines,$a^2 = b^2 + c^2 - 2bc \cos A$,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 $A = 30^{\circ}$,$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$.Thus,$E = 2(\frac{\sqrt{3}}{2}) + 2 = \sqrt{3} + 2$.

With usual notation,in triangle $ABC$,$m \angle A = 30^{\circ}$. Then the value of $\left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right)$ is equal to

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