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

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(B) Let $AD = h$. In $	riangle ADC$,$h = b \sin C$. In $	riangle ADB$,$h = c \sin B$.Given $h = \frac{abc}{b^2 - c^2}$.Using the sine rule,$a = 2R \

Vedclass · Accepted Answer

$113$

Vedclass · Answer

(B) Let $AD = h$. In $	riangle ADC$,$h = b \sin C$. In $	riangle ADB$,$h = c \sin B$.Given $h = \frac{abc}{b^2 - c^2}$.Using the sine rule,$a = 2R \sin A$,$b = 2R \sin B$,$c = 2R \sin C$.Substituting these into the expression for $h$:$h = \frac{(2R \sin A)(2R \sin B)(2R \sin C)}{4R^2 \sin^2 B - 4R^2 \sin^2 C} = \frac{8R^3 \sin A \sin B \sin C}{4R^2 (\sin^2 B - \sin^2 C)} = \frac{2R \sin A \sin B \sin C}{\sin(B-C)\sin(B+C)}$.Since $\sin(B+C) = \sin A$,we have $h = \frac{2R \sin B \sin C}{\sin(B-C)}$.Also,$h = c \sin B = (2R \sin C) \sin B$.Equating the two expressions for $h$:$2R \sin B \sin C = \frac{2R \sin B \sin C}{\sin(B-C)}$.This implies $\sin(B-C) = 1$,so $B - C = 90^o$.Given $C = 23^o$,we have $B - 23^o = 90^o$,which gives $B = 113^o$.

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

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