In Delta ABC, if 8{R^2} = {a^2} + {b^2} + {c^ | vedclass

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

Question

(A) We know that $a = 2R \sin A,$ $b = 2R \sin B,$ and $c = 2R \sin C.$Substituting these in the given equation $8R^2 = a^2 + b^2 + c^2,$$8R^2 = (2R \

Vedclass · Accepted Answer

Right angled

Vedclass · Answer

(A) We know that $a = 2R \sin A,$ $b = 2R \sin B,$ and $c = 2R \sin C.$Substituting these in the given equation $8R^2 = a^2 + b^2 + c^2,$$8R^2 = (2R \sin A)^2 + (2R \sin B)^2 + (2R \sin C)^2$$8R^2 = 4R^2(\sin^2 A + \sin^2 B + \sin^2 C)$$2 = \sin^2 A + \sin^2 B + \sin^2 C$Using $\sin^2 	heta = 1 - \cos^2 	heta,$$2 = (1 - \cos^2 A) + (1 - \cos^2 B) + \sin^2 C$$2 = 2 - \cos^2 A - \cos^2 B + \sin^2 C$$\cos^2 A + \cos^2 B = \sin^2 C = 1 - \cos^2 C$$\cos^2 A + \cos^2 B + \cos^2 C = 1$Using the identity $\cos^2 A + \cos^2 B + \cos^2 C = 1 - 2 \cos A \cos B \cos C,$$1 = 1 - 2 \cos A \cos B \cos C$$2 \cos A \cos B \cos C = 0$This implies $\cos A = 0$ or $\cos B = 0$ or $\cos C = 0.$Therefore,$A = \frac{\pi}{2}$ or $B = \frac{\pi}{2}$ or $C = \frac{\pi}{2}.$Thus,the triangle is a right-angled triangle.

In $\Delta ABC,$ if $8{R^2} = {a^2} + {b^2} + {c^2},$ then the triangle is

Similar Questions

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

In any $\triangle ABC$,$a(b \cos C - c \cos B)$ equals

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

In triangle $ABC$,if $\angle A = 45^\circ$ and $\angle B = 75^\circ$,then $a + c\sqrt{2} = $

In a triangle $ABC$, if $r_1=2 r_2=3 r_3$, then $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module