In any triangle A B C, ((b2-c2)/a2) is equal to

Question

In any triangle A B C, ((b2-c2)/a2) is equal to (A) ( sin (B+C)/ sin (B-C)) (B) ( sin (B-C)/ sin (B+C)) (C)  sin (B+C) (D)  sin (B-C)

Tardigrade Admin · Accepted Answer

∵ ((b2-c2)/a2) =(K2 sin 2 B-K2 sin 2 C/K2 sin 2 A) = (( sin 2 B- sin 2 C)/ sin 2 A)=( sin (B+C) ⋅ sin (B-C)/ sin 2(B+C)) beginBmatrix∵ (A+B+C)=π ⇒ A=π-(B+C) ∴ sin A= sin [π-(B+C)]= sin (B+C) endBmatrix =( sin (B-C)/ sin (B+C)) Hence, ( sin (B-C)/ sin (B+C))=((b2-c2)/a2)

Q. In any $△ A BC, \frac{( b ^{2} - c ^{2} )}{a ^{2}}$ is equal to

$\frac{s i n ( B + C )}{s i n ( B - C )}$

$\frac{s i n ( B - C )}{s i n ( B + C )}$

$sin (B + C)$

$sin (B - C)$

Q. In any △ABC,a2(b2−c2)​ is equal to

sin(B−C)sin(B+C)​

sin(B+C)sin(B−C)​

sin(B+C)

sin(B−C)

∵a2(b2−c2)​=K2sin2AK2sin2B−K2sin2C​ =sin2A(sin2B−sin2C)​=sin2(B+C)sin(B+C)⋅sin(B−C)​ {∵∴​(A+B+C)=π⇒A=π−(B+C)sinA=sin[π−(B+C)]=sin(B+C)​} =sin(B+C)sin(B−C)​ Hence, sin(B+C)sin(B−C)​=a2(b2−c2)​

Q. In any $△ A BC, \frac{( b ^{2} - c ^{2} )}{a ^{2}}$ is equal to

$\frac{s i n ( B + C )}{s i n ( B - C )}$

$\frac{s i n ( B - C )}{s i n ( B + C )}$

$sin (B + C)$

$sin (B - C)$