Question

in a right angle triangle ABC angle is equal to 90 degree and 5 sin square B + 7 cos square c + 4 upon 3 + 8 10 square 60 is equal to 7 upon 27 if AC is equal to 3 find the perimeter of triangle ABC​

Answer 1

Answer:

14.2

Step-by-step explanation:

in a right angle triangle ABC angle is equal to 90 degree and 5 sin square B + 7 cos square c + 4 upon 3 + 8 10 square 60 is equal to 7 upon 27 if AC is equal to 3 find the perimeter of triangle ABC​

$\frac{5Sin^2B + 7Cos^2C + 4}{3 + 8 Tan^260} = \frac{7}{27}$

Tan 60 = √3

Tan²60 = 3

Cos C = SinB as B+C = 90

=> $\frac{5Sin^2B + 7Sin^2B + 4}{3 + (8\times 3)} = \frac{7}{27}$

=>  $\frac{12Sin^2B + 4}{27} = \frac{7}{27}$

=> 12 Sin²B + 4 = 7

=> 12 Sin²B = 3

=> Sin²B = 3/12

=> Sin²B = 1/4

=> Sin B = 1/2

SinB = AC/BC

1/2 = 3/BC

=> BC = 6

AB² = BC² - AC²

=> AB² = 6² - 3²

=> AB² = 36 - 9

=> AB² = 27

=> AB = 3√3

=> AB = 5.2

Perimeter = AB + BC + AC

= 5.2 + 6 + 3

= 14.2

Answer 2

Step-by-step explanation:

this answer is right follow me