Question-72434

Question Number 72434 by aliesam last updated on 28/Oct/19

Answered by mind is power last updated on 28/Oct/19

\frac{c}{\sin (x)} = \frac{d}{\sin (y)} = \frac{BD}{\sin (40)}

{BD}^{2} = a^{2} + b^{2} - 2 abcos (40)

\frac{\sin (y) BD}{\sin (40)} = d, c = \frac{\sin (x) . bD}{\sin (40)}

{(a + b)}^{2} ⩾ 3 (ab + cd) \Leftrightarrow a^{2} + b^{2} - ab ⩾ 3 cd

(a^{2} + b^{2} - 2 abcos (40)) = {BD}^{2} ⩽ a^{2} + b^{2} - ab

2 \cos (40) ⩾ 1

cd = \frac{{BD}^{2} \sin^{z} (x) \sin^{2} (y)}{\sin^{2} (40)}

3 cd = \frac{{3 BD}^{2} \sin^{2} (x) \sin^{2} (y)}{\sin^{2} (40)} ⩽ {BD}^{2}

\Leftrightarrow {3 \sin}^{2} (x) \sin^{2} (y) ⩽ \sin^{2} (40)

x + y = 40

\Rightarrow x or y ⩽ 20

\sin (x) ⩽ \sin (20) ⩽ \sin (30) = \frac{1}{2} ⩽ \frac{1}{\sqrt{3}}

\sin (y) ⩽ \sin (40)

\Rightarrow {3 \sin}^{2} (x) \sin^{2} (y) ⩽ 3 . {(\frac{1}{\sqrt{3}})}^{2} . \sin^{2} (40) = \sin^{2} (40)

\Rightarrow 3 cd ⩽ {DB}^{2} ⩽ a^{2} - ab + b^{2} \Leftrightarrow 3 (cd + ab) ⩽ {(a + b)}^{2}

Commented by aliesam last updated on 28/Oct/19

g o d b l e s s y o u s i r . n i c e s o l u t i o n

Commented by mind is power last updated on 28/Oct/19

y^{'} re welcom