Question-154736

Question Number 154736 by mathdanisur last updated on 21/Sep/21

Commented by Tawa11 last updated on 21/Sep/21

Weldone sir .

Answered by mr W last updated on 21/Sep/21

Commented by mr W last updated on 21/Sep/21

\frac{16}{\tan θ} + 16 + 7 + \frac{7}{\tan (\frac{π}{2} - \frac{θ}{2})} = \frac{16}{\sin θ} + \frac{16}{\cos θ}

23 + 7 \tan \frac{θ}{2} = 16 (\frac{1}{\sin θ} + \frac{1}{\cos θ} - \frac{1}{\tan θ})

l e t t = \tan \frac{θ}{2}

23 + 7 t = 16 (\frac{1 + t^{2}}{2 t} + \frac{1 + t^{2}}{1 - t^{2}} - \frac{1 - t^{2}}{2 t})

23 - 9 t = \frac{16 (1 + t^{2})}{1 - t^{2}}

9 t^{3} - 39 t^{2} - 9 t + 7 = 0

(3 t - 1) (3 t^{2} - 12 t - 7) = 0

t = \frac{1}{3} ✓

t = 2 \pm \frac{\sqrt{51}}{3} (r e j e c t e d, s i n c e 0 < \tan \frac{θ}{2} < 1)

\tan \frac{θ}{2} = \frac{1}{3}

\tan θ = \frac{2 \times \frac{1}{3}}{1 - {(\frac{1}{3})}^{2}} = \frac{3}{4}

A_{b l u e} = \frac{16}{2} \times 16 \tan θ + \frac{7}{2} \times \frac{7}{\tan \frac{θ}{2}}

A_{b l u e} = 128 \times \frac{3}{4} + \frac{49}{2 \times \frac{1}{3}}

A_{b l u e} = \frac{339}{2} = 169.5

Commented by mathdanisur last updated on 21/Sep/21

Creativ solution S er, thank you