In a Delta ABC,angle B = frac{pi}{3},angle C | vedclass

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

Question

(C) Given: $\angle B = \frac{\pi}{3}$,$\angle C = \frac{\pi}{4}$ and $\frac{BD}{DC} = \frac{1}{3}$.In $\Delta ABD$,by the Law of Sines:$\frac{BD}{\sin

Vedclass · Accepted Answer

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

Vedclass · Answer

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

In a $\Delta ABC$,$\angle B = \frac{\pi}{3}$,$\angle C = \frac{\pi}{4}$ and $D$ divides $BC$ internally in the ratio $1:3$,then $\frac{\sin \angle BAD}{\sin \angle CAD}$ is equal to

Similar Questions

If $\tan \theta = \frac{\cos 15^{\circ} + \sin 15^{\circ}}{\cos 15^{\circ} - \sin 15^{\circ}},$ then $\theta =$

If $\theta$ lies in the second quadrant,then the value of $\sqrt{\frac{1 - \sin \theta}{1 + \sin \theta}} + \sqrt{\frac{1 + \sin \theta}{1 - \sin \theta}}$ is:

The equation $(a + b)^2 = 4ab \sin^2 \theta$ is possible only when

If $\sin \alpha = -\frac{3}{5},$ where $\pi < \alpha < \frac{3\pi}{2},$ then $\cos \frac{\alpha}{2} = $

If $x = \sec \phi - \tan \phi$ and $y = \csc \phi + \cot \phi$,then which of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module