Question

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Trigonometry question is given in the attachment.



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Answer 1

HELLO DEAR,

GIVEN that:-
<A = 9/°

<A + < B + <C = 180°

=> <B + <C = 90°

=> <B = (90°- <C)

$\frac{5 {sin}^{2}b + 7 {cos}^{2}c + 4 }{3 + 8 {tan}^{2}60 } = \frac{7}{27} \\ \\ = > \frac{5 {sin}^{2}(90 - c) + 7 {cos}^{2}c + 4 }{3 + 8 {tan}^{2}60 } = \frac{7}{27} \\ \\ = > \frac{5 {cos}^{2}c + 7 {cos}^{2}c + 4 }{3 + 8{ (\ \sqrt{3} })^{2}} = \frac{7}{27} \\ \\ = > \frac{12 {cos}^{2}c + 4 }{3 + 24} = \frac{7}{27} \\ \\ = > 12 {cos}^{2} c + 4 = \frac{7 \times 27}{27} \\ \\ = > 12 {cos}^{2} c = 7 - 4 \\ \\ = > 12 {cos}^{2} = 3 \\ \\ = > {cos}^{2} c = \frac{1}{4} \\ \\ = > cosc = \frac{1}{ 2 }$

cos C = 1/2 = AC/BC

=> 1/2 = 3/BC

=> BC = 6

IN∆ABC ,<A = 90°

BC² = AC² + AB²

=> 6² = 3² + AB²

=> AB² = 36 - 9

=> AB² = 27

=> AB = 3√3

AND,

we know that:-

perimeter of triangle = sum of all sides

=> 3√3 + 6 + 3

=> 3√3 + 9

=> 3√3(1 + 3)
I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Solution is in the attachment.........