In triangle ABC,m angle B = frac{pi}{3} and m | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) In $	riangle ABD$,by the Law of Sines:$\frac{\sin(\angle BAD)}{BD} = \frac{\sin(\angle B)}{AD}$$\Rightarrow \frac{\sin(\angle BAD)}{x} = \frac{\s

Vedclass · Accepted Answer

$\frac{1}{\sqrt{6}}$

Vedclass · Answer

(C) In $	riangle ABD$,by the Law of Sines:$\frac{\sin(\angle BAD)}{BD} = \frac{\sin(\angle B)}{AD}$$\Rightarrow \frac{\sin(\angle BAD)}{x} = \frac{\sin(\frac{\pi}{3})}{AD} = \frac{\sqrt{3}/2}{AD}$$\Rightarrow AD = \frac{\sqrt{3}x}{2 \sin(\angle BAD)} \quad \dots (i)$In $	riangle ADC$,by the Law of Sines:$\frac{\sin(\angle CAD)}{DC} = \frac{\sin(\angle C)}{AD}$$\Rightarrow \frac{\sin(\angle CAD)}{3x} = \frac{\sin(\frac{\pi}{4})}{AD} = \frac{1/\sqrt{2}}{AD}$$\Rightarrow AD = \frac{3x}{\sqrt{2} \sin(\angle CAD)} \quad \dots (ii)$Equating $(i)$ and $(ii)$:$\frac{\sqrt{3}x}{2 \sin(\angle BAD)} = \frac{3x}{\sqrt{2} \sin(\angle CAD)}$$\Rightarrow \frac{\sin(\angle BAD)}{\sin(\angle CAD)} = \frac{\sqrt{3} \cdot \sqrt{2}}{2 \cdot 3} = \frac{\sqrt{6}}{6} = \frac{1}{\sqrt{6}}$

In $\triangle ABC$,$m \angle B = \frac{\pi}{3}$ and $m \angle C = \frac{\pi}{4}$. Let point $D$ divide $BC$ internally in the ratio $1:3$. Then,the value of $\frac{\sin(\angle BAD)}{\sin(\angle CAD)}$ is

Similar Questions

In a triangle $ABC$,if $r_1 r_2 + r_3 r = 35$,$r_2 r_3 + r r_1 = 63$,and $r_3 r_1 + r r_2 = 45$,then $2s =$

In $\triangle ABC$,$a=6 \text{ cm}$,$b=10 \text{ cm}$ and $c=14 \text{ cm}$. Then,the sum of the acute angles of the triangle is (in $^{\circ}$)

In a $\triangle ABC$,if $a=26, b=30$,and $\cos C=\frac{63}{65}$,then $c=$

If one side of a triangle is double the other and the angles opposite to these sides differ by $60^{\circ}$,then the triangle is

With usual notations in $\triangle ABC$,if $\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}$,then $a^2, b^2, c^2$ are in

Vedclass Test Series

Exam Paper Generator

Online Exam Module