Question

In right angled Triangle ABC, angle A = 90° and 5sin ^2 B + 7cos^2 +4 / 3 + 8 tan^2 60 ) = 7 /24
If AC = 3 find perimeter of triangle ABC

Answer 1

Answer:Perimeter = 6 + 3 + 3 root 3 3(3+ root 3)

Step-by-step explanation:

A+B+C=180 B+C+90=180 B+C=90 5sin^2 B+ 7cos^2 C +4/3 + 8tan^2 60=7/27 5sin^2(90-C) +7cos^2 C +4/3 + 8tan^2 60=7/27 5cos^2 C +7cos^2 C +4/3 + 8tan^2 60=7/27 12cos^2 C + 4/ 3 + 8*3 =7/27 12cos^2 C + 4/ 3 + 24 =7/27 12cos^2 C + 4/27 =7/27 12cos^2 C + 4 =7 12cos^2 C =3 Cos^2 C= 1/4 Cos C=1/2 C=60 A=30 Cos60=AC/BC BC=6 sin60= AB/BC AB=3 root 3 Perimeter = 6 + 3 + 3 root 3 3(3+ root 3)