If triangle ABC is a non-isosceles triangle a | vedclass

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

Question

(A) Given that in $	riangle ABC$,$\angle C = 90^{\circ}$.Since $\angle A + \angle B + \angle C = 180^{\circ}$,we have $\angle A + \angle B = 90^{\cir

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given that in $	riangle ABC$,$\angle C = 90^{\circ}$.Since $\angle A + \angle B + \angle C = 180^{\circ}$,we have $\angle A + \angle B = 90^{\circ}$.Using the sine rule,$a = k \sin A$ and $b = k \sin B$.Substituting these into the expression:$\frac{a^2+b^2}{a^2-b^2} \sin(A-B) = \frac{k^2 \sin^2 A + k^2 \sin^2 B}{k^2 \sin^2 A - k^2 \sin^2 B} \sin(A-B) = \frac{\sin^2 A + \sin^2 B}{\sin^2 A - \sin^2 B} \sin(A-B)$.Using the identity $\sin^2 A - \sin^2 B = \sin(A+B) \sin(A-B)$:$= \frac{\sin^2 A + \sin^2 B}{\sin(A+B) \sin(A-B)} \sin(A-B) = \frac{\sin^2 A + \sin^2 B}{\sin(A+B)}$.Since $A+B = 90^{\circ}$,$\sin(A+B) = 1$ and $B = 90^{\circ}-A$,so $\sin B = \cos A$.$= \frac{\sin^2 A + \cos^2 A}{1} = 1$.

If $\triangle ABC$ is a non-isosceles triangle and $\angle C = 90^{\circ}$,then $\frac{a^2+b^2}{a^2-b^2} \sin(A-B) = $

Similar Questions

If in a triangle $ABC$,$\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13}$,then $\cos A$ is equal to

In a $\triangle ABC$,if $a \cos A = b \cos B$,where $a \neq b$,then $\triangle ABC$ is

In $\Delta ABC$,if $2s = a + b + c$,then the value of $\frac{s(s - a)}{bc} - \frac{(s - b)(s - c)}{bc} = $

In a $\triangle ABC$,if $\angle A = 60^{\circ}$,then $(a+b+c)(b+c-a) =$

In a $\triangle ABC$,if $a^4+b^4+c^4=2b^2c^2+2a^2b^2$,then $B=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module