If 4 sin A=4 sin B=3 sin C in a triangle ABC, then cos C is equal to

Question

If 4 sin A=4 sin B=3 sin C in a triangle ABC, then cos C is equal to (A)  (1/3)  (B)  (1/9)  (C)  (1/27)  (D)  (1/18)

Tardigrade Admin · Accepted Answer

Given, 4 sin A=4 sin B=3 sin C or ( sin A/1/4)=( sin B/1/4)=( sin C/1/3) or ( sin A/3)=( sin B/3)=( sin C/4) Here, a=3k,b=3k and c=4k ∴ cos C=(a2+b2-c2/2ab) =(9k2+9k2-16k2/2× 3k× 3k)=(1/9)

Q. If $4 sin A = 4 sin B = 3 sin C$ in a triangle $A BC$ , then $cos C$ is equal to

$\frac{1}{3}$

$\frac{1}{9}$

$\frac{1}{27}$

$\frac{1}{18}$

Q. If 4sinA=4sinB=3sinC in a triangle ABC, then cosC is equal to

31​

91​

271​

181​

Given, 4sinA=4sinB=3sinC or 1/4sinA​=1/4sinB​=1/3sinC​ or 3sinA​=3sinB​=4sinC​ Here, a=3k,b=3k and c=4k ∴ cosC=2aba2+b2−c2​ =2×3k×3k9k2+9k2−16k2​=91​

Q. If $4 sin A = 4 sin B = 3 sin C$ in a triangle $A BC$ , then $cos C$ is equal to

$\frac{1}{3}$

$\frac{1}{9}$

$\frac{1}{27}$

$\frac{1}{18}$