Let a, b and c denote the lengths of sides BC | vedclass

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

Question

(B) Given $BC = a, CA = b$ and $AB = c$.To find $a: b: c$.Since the sum of angles in a triangle is $180^{\circ}$,we have:$\angle ACB = 180^{\circ} - (

Vedclass · Accepted Answer

$1: \sqrt{3}: 2$

Vedclass · Answer

(B) Given $BC = a, CA = b$ and $AB = c$.To find $a: b: c$.Since the sum of angles in a triangle is $180^{\circ}$,we have:$\angle ACB = 180^{\circ} - (\angle BAC + \angle ABC)$$\angle ACB = 180^{\circ} - (30^{\circ} + 60^{\circ}) = 180^{\circ} - 90^{\circ} = 90^{\circ}$.By using the sine rule,we have:$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$\frac{a}{\sin 30^{\circ}} = \frac{b}{\sin 60^{\circ}} = \frac{c}{\sin 90^{\circ}}$Substituting the values of the sine functions:$\frac{a}{1/2} = \frac{b}{\sqrt{3}/2} = \frac{c}{1}$Multiplying by $1/2$,we get:$a : b : c = \frac{1}{2} : \frac{\sqrt{3}}{2} : 1$Multiplying the ratio by $2$,we get:$a : b : c = 1 : \sqrt{3} : 2$.

Let $a, b$ and $c$ denote the lengths of sides $BC, CA$ and $AB$ of $\triangle ABC$. In $\triangle ABC$,$\angle BAC = 30^{\circ}$ and $\angle ABC = 60^{\circ}$. Then $a: b: c$ is

Similar Questions

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

In a $\Delta ABC$,$\frac{a}{b} = 2 + \sqrt{3}$ and $\angle C = 60^\circ$. Then the ordered pair $(\angle A, \angle B)$ is equal to

If the angles of a triangle are in the ratio $4:1:1$,then the ratio of the longest side to its perimeter is

For the triangle $ABC$,with usual notations,if the angles $A, B, C$ are in $A.P.$ and $m \angle A = 30^{\circ}, c = 3$,then the values of $a$ and $b$ are respectively

The inradius of $\Delta ABC$ is $3$ units and the exradius opposite to vertex $A$ is $4$ units. Then,the length of the altitude from vertex $A$ is ............. $units$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module