If 4 times the area of a triangle ABC is c2-(a-b)2, then sin C

Question

If 4 times the area of a triangle ABC is c2-(a-b)2, then sin C (A) (√3/2) (B) (1/√2) (C) (1/2) (D) 1

Tardigrade Admin · Accepted Answer

4 Δ=c2-(a-b)2 4 ⋅ (1/2) a b sin C=c2-a2-b2+2 a b sin C=1-(a2+b2-c2/2 a b) ⇒ sin C=1- cos C ⇒ sin C+ cos C=1 ⇒ √2 sin (C+(π/4))=1 C+(π/4)=(3 π/4) ∴ C=(π/2)

Q. If 4 times the area of a $△ A BC$ is $c^{2} - (a - b)^{2}$ , then $sin C$

$\frac{3}{2}$

$\frac{1}{2}$

$\frac{1}{2}$

1

Q. If 4 times the area of a △ABC is c2−(a−b)2, then sinC

23​​

2​1​

21​

1

4Δ=c2−(a−b)2 4⋅21​absinC=c2−a2−b2+2ab sinC=1−2aba2+b2−c2​ ⇒sinC=1−cosC⇒sinC+cosC=1 ⇒2​sin(C+4π​)=1 C+4π​=43π​ ∴C=2π​

Q. If 4 times the area of a $△ A BC$ is $c^{2} - (a - b)^{2}$ , then $sin C$

$\frac{3}{2}$

$\frac{1}{2}$

$\frac{1}{2}$