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

(C) Let the angles be $x, 2x,$ and $7x.$ Since the sum of angles in a triangle is $180^{\circ},$ we have $x + 2x + 7x = 180^{\circ}$ $\Rightarrow 10x

Vedclass · Accepted Answer

$(\sqrt{5} + 1):(\sqrt{5} - 1)$

Vedclass · Answer

(C) Let the angles be $x, 2x,$ and $7x.$ Since the sum of angles in a triangle is $180^{\circ},$ we have $x + 2x + 7x = 180^{\circ}$ $\Rightarrow 10x = 180^{\circ}$ $\Rightarrow x = 18^{\circ}.$ Thus,the angles are $18^{\circ}, 36^{\circ},$ and $126^{\circ}.$ By the Sine Rule,the sides are proportional to the sines of the opposite angles. Let the sides be $a, b, c$ opposite to $18^{\circ}, 36^{\circ}, 126^{\circ}$ respectively. The greatest side is $c$ and the least side is $a.$ The ratio is $\frac{c}{a} = \frac{\sin(126^{\circ})}{\sin(18^{\circ})}.$ Since $\sin(126^{\circ}) = \sin(180^{\circ} - 54^{\circ}) = \sin(54^{\circ}) = \cos(36^{\circ}) = \frac{\sqrt{5} + 1}{4}$ and $\sin(18^{\circ}) = \frac{\sqrt{5} - 1}{4}.$ Therefore,the ratio is $\frac{(\sqrt{5} + 1)/4}{(\sqrt{5} - 1)/4} = \frac{\sqrt{5} + 1}{\sqrt{5} - 1}.$

If the angles of a triangle are in the ratio $1 : 2 : 7,$ then the ratio of its greatest side to the least side is

Similar Questions

In a $\Delta ABC$,$a, b, A$ are given and $c_1, c_2$ are two values of the third side $c$. The sum of the areas of the two triangles with sides $(a, b, c_1)$ and $(a, b, c_2)$ is

In a triangle $ABC$,with usual notations,if $m \angle A = 60^{\circ}$,$b = 8$,$a = 6$,and $B = \sin^{-1} x$,then $x$ has the value:

In a $\triangle ABC$,if $3a = b + c$,then $\cot \frac{B}{2} \cot \frac{C}{2}$ is equal to :

If $A, B, C$ are the angles of a triangle with $\tan \frac{A}{2}=\frac{1}{3}$ and $\tan \frac{B}{2}=\frac{2}{3}$,then the value of $\tan \frac{C}{2}$ is:

In $\triangle ABC$,if $r=3$ and $R=5$,then $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module