In a triangle ABC,with usual notation,angle B | vedclass

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Question

(A) Let $AB = c$,$AC = b$,and $BC = a$. Let $\angle BAD = \alpha$ and $\angle CAD = \beta$. In $	riangle ABD$,by the sine rule: $\frac{BD}{\sin \alph

Vedclass · Accepted Answer

$1/\sqrt{6}$

Vedclass · Answer

(A) Let $AB = c$,$AC = b$,and $BC = a$. Let $\angle BAD = \alpha$ and $\angle CAD = \beta$. In $	riangle ABD$,by the sine rule: $\frac{BD}{\sin \alpha} = \frac{c}{\sin \angle ADB} \implies \sin \alpha = \frac{BD \sin \angle ADB}{c}$. In $	riangle ACD$,by the sine rule: $\frac{CD}{\sin \beta} = \frac{b}{\sin \angle ADC} \implies \sin \beta = \frac{CD \sin \angle ADC}{c}$. Since $\angle ADB + \angle ADC = \pi$,$\sin \angle ADB = \sin \angle ADC$. Thus,$\frac{\sin \alpha}{\sin \beta} = \frac{BD}{CD} \cdot \frac{b}{c}$. Given $BD:CD = 1:3$,so $BD/CD = 1/3$. By the sine rule in $	riangle ABC$,$\frac{b}{\sin B} = \frac{c}{\sin C} \implies \frac{b}{c} = \frac{\sin(\pi/3)}{\sin(\pi/4)} = \frac{\sqrt{3}/2}{1/\sqrt{2}} = \frac{\sqrt{3}}{\sqrt{2}}$. Therefore,$\frac{\sin \alpha}{\sin \beta} = \frac{1}{3} \cdot \frac{\sqrt{3}}{\sqrt{2}} = \frac{1}{\sqrt{3} \cdot \sqrt{2}} = \frac{1}{\sqrt{6}}$.

In a triangle $ABC$,with usual notation,$\angle B = \pi/3$ and $\angle C = \pi/4$. If $D$ divides $BC$ internally in the ratio $1:3$,find the value of $\frac{\sin \angle BAD}{\sin \angle CAD}$.

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