In triangle ABC,if sin^2 B = sin C and 3 cos^ | vedclass

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

Question

(D) Given $3 \cos^2 B = 2 \cos^2 C$. Substituting $\cos^2 	heta = 1 - \sin^2 	heta$,we get $3(1 - \sin^2 B) = 2(1 - \sin^2 C)$. Since $\sin^2 B = \s

Vedclass · Accepted Answer

a scalene triangle

Vedclass · Answer

(D) Given $3 \cos^2 B = 2 \cos^2 C$. Substituting $\cos^2 	heta = 1 - \sin^2 	heta$,we get $3(1 - \sin^2 B) = 2(1 - \sin^2 C)$. Since $\sin^2 B = \sin C$,we substitute this into the equation: $3(1 - \sin C) = 2(1 - \sin^2 C)$. $3 - 3 \sin C = 2 - 2 \sin^2 C$. $2 \sin^2 C - 3 \sin C + 1 = 0$. $(2 \sin C - 1)(\sin C - 1) = 0$. Thus,$\sin C = \frac{1}{2}$ or $\sin C = 1$. If $\sin C = 1$,then $C = \frac{\pi}{2}$,which implies $\sin^2 B = \sin C = 1$,so $B = \frac{\pi}{2}$. This is impossible in a triangle as $B+C = \pi$. Therefore,$\sin C = \frac{1}{2}$,so $C = \frac{\pi}{6}$ or $C = \frac{5\pi}{6}$. If $C = \frac{\pi}{6}$,then $\sin^2 B = \sin(\frac{\pi}{6}) = \frac{1}{2}$,so $\sin B = \frac{1}{\sqrt{2}}$,which means $B = \frac{\pi}{4}$ or $B = \frac{3\pi}{4}$. If $B = \frac{\pi}{4}$ and $C = \frac{\pi}{6}$,then $A = \pi - (\frac{\pi}{4} + \frac{\pi}{6}) = \pi - \frac{5\pi}{12} = \frac{7\pi}{12}$. Since all angles $A, B, C$ are distinct,$	riangle ABC$ is a scalene triangle.

In $\triangle ABC$,if $\sin^2 B = \sin C$ and $3 \cos^2 B = 2 \cos^2 C$,then $\triangle ABC$ is

Similar Questions

The product of the arithmetic mean of the lengths of the sides of a triangle and the harmonic mean of the lengths of the altitudes of the triangle is equal to: [where $\Delta$ is the area of the triangle $ABC$]

The set of solutions of the system of equations $x+y = \frac{2 \pi}{3}$ and $\cos x + \cos y = \frac{3}{2}$,where $x, y$ are real,is

In triangle $ABC$,if $A=45^{\circ}$,$C=75^{\circ}$ and $R=\sqrt{2}$,then $r=$

In $\triangle ABC$,if $\cos ^2 A + \cos ^2 B + \cos ^2 C = 1$,then $\triangle ABC$ is

The value of $\theta$,satisfying both the equations $\cos \theta = \frac{1}{\sqrt{2}}$ and $\tan \theta = -1$ in the interval $[0, 2\pi]$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module