If one side of a triangle is double the other | vedclass

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

Question

(B) In $	riangle ABC$,by the sine rule,$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$.Given $a = 2b$ and $A - B = 60^{\circ}$,so $A = 60^{\

Vedclass · Accepted Answer

right angled

Vedclass · Answer

(B) In $	riangle ABC$,by the sine rule,$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$.Given $a = 2b$ and $A - B = 60^{\circ}$,so $A = 60^{\circ} + B$.Substituting these into the sine rule:$\frac{\sin(60^{\circ} + B)}{2b} = \frac{\sin B}{b}$$\sin(60^{\circ} + B) = 2 \sin B$$\sin 60^{\circ} \cos B + \cos 60^{\circ} \sin B = 2 \sin B$$\frac{\sqrt{3}}{2} \cos B + \frac{1}{2} \sin B = 2 \sin B$$\frac{\sqrt{3}}{2} \cos B = \frac{3}{2} \sin B$$	an B = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$.Thus,$B = 30^{\circ}$.Then $A = 60^{\circ} + 30^{\circ} = 90^{\circ}$.Since one angle is $90^{\circ}$,the triangle is right angled.

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

Similar Questions

In a $\triangle ABC$,if $a+3b=3c$,then $\sin \frac{A}{2} =$

In $\Delta ABC,$ $1 - \tan \frac{A}{2}\tan \frac{B}{2} = $

In a $\triangle ABC$,$b:c = \sqrt{3}:\sqrt{2}$ and the angles $A, B, C$ are in $AP$,then $\angle A = $ (in $^{\circ}$)

In a $\Delta ABC,$ if $(\sin A + \sin B + \sin C)(\sin A + \sin B - \sin C) = 3\sin A\sin B,$ then the angle $C$ is equal to

In $\triangle ABC$,if $a, b, c$ are in arithmetic progression and $A=2C$,then $b:c=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module