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

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

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 $\frac{\sin \angle BAD}{\sin \angle CAD}$.

Similar Questions

In $\triangle ABC$, with usual notations, if $\cos A = \frac{1}{2}$, $a = 3$, and $\angle B = \frac{\pi}{6}$, then the values of $b$ and $c$ are:

In a triangle $ABC$, if $r_1=2 r_2=3 r_3$, then $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=$

For the triangle shown in the figure,side $b = 10 \, m$,and angles $\angle C$ and $\angle A$ are equal. Find the lengths of sides $a$ and $c$.

The perimeter of $\Delta ABC$ is $6$ times the $A.M.$ of the sines of its angles. If $a = 1$,then $\angle A = $

In a triangle $ABC$,if $A = 2B$ and the sides opposite to the angles $A, B, C$ are $\alpha + 1, \alpha - 1$ and $\alpha$ respectively,then $\alpha =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module