If the sides of a triangle are in the ratio s | 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 $a = \sqrt{3}k$,$b = \sqrt{5}k$,and $c = \sqrt{8+\sqrt{15}}k$. Since $c$ is the largest side,the largest angle $\

Vedclass · Accepted Answer

$\frac{2 \pi}{3}$

Vedclass · Answer

(A) Let the sides of the triangle be $a = \sqrt{3}k$,$b = \sqrt{5}k$,and $c = \sqrt{8+\sqrt{15}}k$. Since $c$ is the largest side,the largest angle $	heta$ is opposite to side $c$. Using the Law of Cosines: $\cos 	heta = \frac{a^2 + b^2 - c^2}{2ab}$. Substituting the values: $\cos 	heta = \frac{(\sqrt{3}k)^2 + (\sqrt{5}k)^2 - (\sqrt{8+\sqrt{15}}k)^2}{2(\sqrt{3}k)(\sqrt{5}k)}$. $\cos 	heta = \frac{3k^2 + 5k^2 - (8+\sqrt{15})k^2}{2\sqrt{15}k^2}$. $\cos 	heta = \frac{8k^2 - 8k^2 - \sqrt{15}k^2}{2\sqrt{15}k^2} = \frac{-\sqrt{15}k^2}{2\sqrt{15}k^2} = -\frac{1}{2}$. Since $\cos 	heta = -\frac{1}{2}$,the angle $	heta = \frac{2\pi}{3}$.

If the sides of a triangle are in the ratio $\sqrt{3} : \sqrt{5} : \sqrt{8+\sqrt{15}}$,then the largest angle in that triangle is

Similar Questions

In a $\triangle ABC$,if $r_1=12, r_2=18$ and $r_3=36$,then $b=$

In $\Delta ABC$ with usual notations $a=4, b=3, \angle A=60^{\circ}$,then $c$ is a root of the equation

If in a $\triangle ABC$,$r_1 < r_2 < r_3$,then:

If $\frac{\sin A - \sin C}{\cos C - \cos A} = \cot B$,then $A, B, C$ are in

In a $\triangle ABC$,if $r_1 = 2r_2 = 3r_3$,then the ratio $a : b$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module