In a triangle ABC,AD is the altitude from A. | vedclass

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(C) In $	riangle ABC$,$AD = b \sin C = c \sin B$. Given $AD = \frac{abc}{b^2 - c^2}$,we have $AD(b^2 - c^2) = abc$. Substituting $AD = b \sin C$ and

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$113$

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(C) In $	riangle ABC$,$AD = b \sin C = c \sin B$. Given $AD = \frac{abc}{b^2 - c^2}$,we have $AD(b^2 - c^2) = abc$. Substituting $AD = b \sin C$ and $AD = c \sin B$: $b \sin C (b^2 - c^2) = abc \implies \sin C (b^2 - c^2) = ac$. Using the sine rule $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$: $\sin C (4R^2 \sin^2 B - 4R^2 \sin^2 C) = (2R \sin A)(2R \sin C)$. $\sin^2 B - \sin^2 C = \sin A = \sin(180^\circ - (B+C)) = \sin(B+C)$. $(\sin B - \sin C)(\sin B + \sin C) = \sin B \cos C + \cos B \sin C$. Since $\sin B - \sin C = 2 \sin(\frac{B-C}{2}) \cos(\frac{B+C}{2})$ and $\sin B + \sin C = 2 \sin(\frac{B+C}{2}) \cos(\frac{B-C}{2})$,$2 \sin(\frac{B-C}{2}) \cos(\frac{B+C}{2}) \cdot 2 \sin(\frac{B+C}{2}) \cos(\frac{B-C}{2}) = \sin(B+C)$. $2 \sin(B-C) \cdot 2 \sin(B+C) \cdot \frac{1}{2} = \sin(B+C)$. $2 \sin(B-C) = 1 \implies \sin(B-C) = \frac{1}{2} = \sin 30^\circ$. $B - C = 30^\circ \implies B = 30^\circ + 23^\circ = 53^\circ$ (Not in options). Re-evaluating the condition $AD = \frac{abc}{b^2-c^2}$: $AD = \frac{abc}{b^2-c^2} \implies \frac{1}{AD} = \frac{b^2-c^2}{abc} = \frac{b}{ac} - \frac{c}{ab} = \frac{\sin B}{a \sin C} - \frac{\sin C}{a \sin B} = \frac{\sin^2 B - \sin^2 C}{a \sin B \sin C}$. Since $AD = c \sin B$,$\frac{1}{c \sin B} = \frac{\sin(B-C)\sin(B+C)}{a \sin B \sin C}$. Using $a = \frac{c \sin A}{\sin C} = \frac{c \sin(B+C)}{\sin C}$,we get $\sin(B-C) = 1$. $B - C = 90^\circ \implies B = 90^\circ + 23^\circ = 113^\circ$.

In a triangle $ABC$,$AD$ is the altitude from $A$. Given $b > c$,$\angle C = 23^\circ$ and $AD = \frac{abc}{b^2 - c^2}$,then $\angle B = $ .....$^\circ$

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