Question

Please help mee! thanks x

Answer 1

Step-by-step explanation:

ar of ∆ A = 1/2*x*h = ar of ∆ B = 1/2*1*(x+1)

=> x*h = x + 1

=> x*h - x = 1

=> x(h - 1) = 1

=> x = 1/(h - 1)........1

in ∆ A, sin 30° = h/x

=> 1/2 = h/x

=> 2h = x

=> h = x/2.......2

substituting value of h from equation 1 in equation 2, we get

x = 1/(x/2 - 1)

=> 1/(x - 2)/2

=> 2/(x - 2)

=> x(x-2) = 2

=> x² - 2x - 2 = 0

a= 1, b = -2 , c = -2

by applying quadratic equation,

x = [- b ± ✓(b² - 4ac)] / 2a

=> x = [-(-2) ± √ ((-2)² - 4*1*(-2)] / 2*1

=> x = [2 ± √ (4 +8)] / 2

=> x = (2 ± √12) / 2

=> x = (2 ± 2√3) / 2

=> x = 2(1 ± √3) / 2

=> x = 1 ± √3 where a = 1 and b = 3