The sides of a triangle are in the ratio 1 : | vedclass

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

Question

(A) Let the sides of the triangle be $a = k$,$b = \sqrt{3}k$,and $c = 2k$.Since $a^2 + b^2 = k^2 + (\sqrt{3}k)^2 = k^2 + 3k^2 = 4k^2 = c^2$,the triang

Vedclass · Accepted Answer

$1 : 2 : 3$

Vedclass · Answer

(A) Let the sides of the triangle be $a = k$,$b = \sqrt{3}k$,and $c = 2k$.Since $a^2 + b^2 = k^2 + (\sqrt{3}k)^2 = k^2 + 3k^2 = 4k^2 = c^2$,the triangle is a right-angled triangle with the hypotenuse $c = 2k$.Let the angles opposite to sides $a, b, c$ be $A, B, C$ respectively.Then $C = 90^{\circ}$.Using the sine rule or trigonometric ratios in a right triangle:$\sin A = \frac{a}{c} = \frac{k}{2k} = \frac{1}{2} \implies A = 30^{\circ}$.$\sin B = \frac{b}{c} = \frac{\sqrt{3}k}{2k} = \frac{\sqrt{3}}{2} \implies B = 60^{\circ}$.Thus,the angles are $30^{\circ}, 60^{\circ}, 90^{\circ}$.The ratio of the angles is $30^{\circ} : 60^{\circ} : 90^{\circ} = 1 : 2 : 3$.

The sides of a triangle are in the ratio $1 : \sqrt{3} : 2$. Then the angles are in the ratio

Similar Questions

In $\Delta ABC$,$8 \Delta = (b + c)(bc + 1)$,then the circumradius of $\Delta ABC$ is (where $\Delta$ denotes the area of the triangle and $b, c$ are the lengths of sides $AC$ and $AB$ respectively):

Let $ABC$ be an acute-angled triangle with area $R$. Then,$\sqrt{a^2 b^2-4 R^2}+\sqrt{b^2 c^2-4 R^2}+\sqrt{c^2 a^2-4 R^2} = $

In $\triangle ABC$,$a=6 \text{ cm}$,$b=10 \text{ cm}$ and $c=14 \text{ cm}$. Then,the sum of the acute angles of the triangle is (in $^{\circ}$)

Let $p, q$ and $r$ be the sides opposite to the angles $P, Q$ and $R$ respectively in a $\Delta PQR$. Then, $2pr \sin \left(\frac{P-Q+R}{2}\right)$ equals

The lengths of the sides of a triangle are three consecutive natural numbers and its largest angle is twice the smallest one. Then the lengths of the sides of the triangle (in units) are

Vedclass Test Series

Exam Paper Generator

Online Exam Module