If the area of triangle triangle A B C is a2-(b-c)2 then its circumradius R=

Question

If the area of triangle triangle A B C is a2-(b-c)2 then its circumradius R= (A) a / 16 operatornamecosec2(A / 2) (B) (C / 16) sin 2(C / 2) (C) a / 6 sin 2(A / 2) (D) b / 16 sin 2(B / 2)

Tardigrade Admin · Accepted Answer

Δ=a2-(b-c)2=a2-b2-c2+2 b c cos A=(b2+c2-a2/2 b c) =(2 b c-Δ/2 b c)=1-(Δ/2 b c) =1-(a b c/4 R × 2 b c)=1-(a/8 R) ⇒ 8 R=(a/1- cos A)=(a/2 sin 2(A / 2)) ⇒ R=((a/16)) textcosec 2((A/2))

Q. If the area of triangle $△ A BC$ is $a^{2} - (b - c)^{2}$ then its circumradius $R =$

$a /16 cosec^{2} (A /2)$

$(C /16) sin^{2} (C /2)$

$a /6 sin^{2} (A /2)$

$b /16 sin^{2} (B /2)$

Q. If the area of triangle △ABC is a2−(b−c)2 then its circumradius R=

a/16cosec2(A/2)

(C/16)sin2(C/2)

a/6sin2(A/2)

b/16sin2(B/2)

Δ=a2−(b−c)2=a2−b2−c2+2bc cosA=2bcb2+c2−a2​ =2bc2bc−Δ​=1−2bcΔ​ =1−4R×2bcabc​=1−8Ra​ ⇒8R=1−cosAa​=2sin2(A/2)a​ ⇒R=(16a​)cosec2(2A​)

Q. If the area of triangle $△ A BC$ is $a^{2} - (b - c)^{2}$ then its circumradius $R =$

$a /16 cosec^{2} (A /2)$

$(C /16) sin^{2} (C /2)$

$a /6 sin^{2} (A /2)$

$b /16 sin^{2} (B /2)$