If the angles of a triangle are in the ratio | vedclass

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

Question

(B) Let the angles of the triangle be $x, 2x, 3x$.Since the sum of angles in a triangle is $180^\circ$,we have $x + 2x + 3x = 180^\circ$,which gives $

Vedclass · Accepted Answer

$1:\sqrt{3}:2$

Vedclass · Answer

(B) Let the angles of the triangle be $x, 2x, 3x$.Since the sum of angles in a triangle is $180^\circ$,we have $x + 2x + 3x = 180^\circ$,which gives $6x = 180^\circ$,so $x = 30^\circ$.The angles are $30^\circ, 60^\circ, 90^\circ$.By the Sine Rule,the sides $a, b, c$ are proportional to the sines of their opposite angles:$a:b:c = \sin(30^\circ) : \sin(60^\circ) : \sin(90^\circ)$.Substituting the values,we get $a:b:c = \frac{1}{2} : \frac{\sqrt{3}}{2} : 1$.Multiplying by $2$,we get $a:b:c = 1 : \sqrt{3} : 2$.

If the angles of a triangle are in the ratio $1:2:3$,then their corresponding sides are in the ratio

Similar Questions

In $\triangle ABC$,if $a: b: c = 4: 5: 6$,then $\frac{\cos A + 3 \cos C}{\cos B} = $

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 triangle $ABC$,$l(AB)=\sqrt{23}$ units,$l(BC)=3$ units,$l(CA)=4$ units,then $\frac{\cot A+\cot C}{\cot B}$ is

In $\triangle ABC$, with usual notations, if the sides $a, b$ and $c$ are in the ratio $18 : 17 : 7$, then $\cot \frac{A}{2} : \cot \frac{B}{2} : \cot \frac{C}{2} = $

With usual notations in $\Delta ABC$,$a=3$,$c=2$ and $\sin C=\frac{2}{3}$,then $\angle A=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module