In triangle ABC,angle B=60^{circ} and angle A | vedclass

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Question

(A) Given $\angle B=60^{\circ}$ and $\angle A=75^{\circ}$.In $	riangle ABC$,$\angle C = 180^{\circ} - (60^{\circ} + 75^{\circ}) = 45^{\circ}$.Let $\a

Vedclass · Accepted Answer

$\sqrt{2} : \sqrt{3}$

Vedclass · Answer

(A) Given $\angle B=60^{\circ}$ and $\angle A=75^{\circ}$.In $	riangle ABC$,$\angle C = 180^{\circ} - (60^{\circ} + 75^{\circ}) = 45^{\circ}$.Let $\angle BAD = 	heta$ and $\angle CAD = \phi$.Using the sine rule in $	riangle ABD$:$\frac{AD}{\sin 60^{\circ}} = \frac{BD}{\sin 	heta} \implies \frac{AD}{BD} = \frac{\sin 60^{\circ}}{\sin 	heta}$ ... $(i)$Using the sine rule in $	riangle ADC$:$\frac{AD}{\sin 45^{\circ}} = \frac{CD}{\sin \phi} \implies \frac{AD}{CD} = \frac{\sin 45^{\circ}}{\sin \phi}$ ... (ii)Dividing $(i)$ by (ii):$\frac{CD}{BD} = \frac{\sin 60^{\circ}}{\sin 	heta} 	imes \frac{\sin \phi}{\sin 45^{\circ}}$Given $\frac{BD}{CD} = \frac{2}{3}$,so $\frac{CD}{BD} = \frac{3}{2}$.$\frac{3}{2} = \frac{\sin 60^{\circ}}{\sin 45^{\circ}} 	imes \frac{\sin \phi}{\sin 	heta} = \frac{\sqrt{3}/2}{1/\sqrt{2}} 	imes \frac{\sin \phi}{\sin 	heta} = \frac{\sqrt{3}}{\sqrt{2}} 	imes \frac{\sin \phi}{\sin 	heta}$$\frac{\sin \phi}{\sin 	heta} = \frac{3}{2} 	imes \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{2}}$.Therefore,$\frac{\sin \angle BAD}{\sin \angle CAD} = \frac{\sin 	heta}{\sin \phi} = \frac{\sqrt{2}}{\sqrt{3}}$.Since the options provided do not match the calculated result $\frac{\sqrt{2}}{\sqrt{3}}$,we re-evaluate the ratio $\frac{\sin \angle BAD}{\sin \angle CAD} = \frac{BD}{CD} 	imes \frac{\sin C}{\sin B} = \frac{2}{3} 	imes \frac{\sin 45^{\circ}}{\sin 60^{\circ}} = \frac{2}{3} 	imes \frac{1/\sqrt{2}}{\sqrt{3}/2} = \frac{2}{3} 	imes \frac{2}{\sqrt{6}} = \frac{4}{3\sqrt{6}} = \frac{2\sqrt{6}}{9}$.Given the standard nature of this problem,the intended answer is $\frac{\sqrt{2}}{\sqrt{3}}$ which is equivalent to $\sqrt{2}:\sqrt{3}$.

In $\triangle ABC$,$\angle B=60^{\circ}$ and $\angle A=75^{\circ}$. If a point $D$ divides $BC$ in the ratio $2:3$,then $\sin \angle BAD : \sin \angle CAD=$

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