Question

In triangle ABC is b square + c square upon b square minus c square is equal to sin(B+C) upon sin(B-C) then which triangle ABC is

Answer 1

here's your answer

hope this helps

please Mark as brainlist

you can follow me too

Answer 2

Answer:

Triangle ABC is a right angled isosceles triangle.

Step-by-step explanation:

Given a triangle ABC and the condition

$\dfrac{b^{2} + c^{2} }{b^{2} - c^{2}} =\dfrac{sin(B+C)}{ sin(B-C)}$

Using the componendo and dividendo rule, which says, if we have

$\dfrac{a }{b} =\dfrac{c}{d}$

then $\dfrac{a+b }{a-b} =\dfrac{c+d}{c-d}$.

Applying this to the given condition,

$\dfrac{b^{2} + c^{2} +(b^{2} - c^{2})}{b^{2} + c^{2}-(b^{2} - c^{2})} =\dfrac{sin(B+C)+sin(B-C)}{ sin(B+C)-sin(B-C)}$

Using the trigonometric relations,

$2sinAcosB={sin(A+B)+sin(A-B)}$

and $2cosAsinB= sin(A+B)-sin(A-B)}$

$\implies \dfrac{b^{2} +b^{2}}{c^{2} + c^{2}} =\dfrac{2sinBcosC}{2cosBsinC}$

$\implies \dfrac{b^{2}}{c^{2}} =\dfrac{sinBcosC}{cosBsinC}$

Applying the sin rule, for a triangle ABC

$\dfrac{a}{sinA }-\dfrac{b}{sinB }=\dfrac{c}{sinC }=\dfrac{1}{K }$

$\dfrac{b^{2}}{c^{2}} =\dfrac{bKcosC}{cosBcK}$

$\implies \dfrac{b}{c} =\dfrac{cosC}{cosB}$

Using a cosine rule, for an angle A in a triangle

$cosA=\dfrac{b^{2}+ c^{2}-a^{2} }{2bc}$

$\dfrac{b}{c} =\dfrac{\dfrac{b^{2}- c^{2}+a^{2} }{2ab}}{\dfrac{-b^{2}+ c^{2}+a^{2} }{2ac}}$

$\implies \dfrac{b}{c} ={\dfrac{(b^{2}- c^{2}+a^{2} )c}{b({-b^{2}+ c^{2}+a^{2} )}}$

$\implies {b^{2} ({-b^{2}+ c^{2}+a^{2} )}=(b^{2}- c^{2}+a^{2} )c^{2}$

$\implies {b^{2} ({-b^{2}+ a^{2} )}=(- c^{2}+a^{2} )c^{2}$

$\implies {b^{4}- c^{4}=b^{2}a^{2}-a^{2} c^{2}$

$\implies (b^{2}+ c^{2})(b^{2}- c^{2})-a^{2} (b^{2}- c^{2})=0$

$\implies (b^{2}- c^{2})((b^{2}+ c^{2})-a^{2}) =0$  

$\implies (b^{2}+ c^{2})-a^{2}=0 ~\mbox{or}~ b^{2}- c^{2}=0$

$\implies b^{2}+ c^{2}=a^{2} ~\mbox{or}~ b^{2}= c^{2}$

Two sides are equal, and square of a side is equal to the sum of the squares of the other two sides.

Therefore, triangle ABC is a right angled isosceles triangle.