Question-186079

Question Number 186079 by TUN last updated on 31/Jan/23

Answered by a.lgnaoui last updated on 01/Feb/23

△ A B D e t △ A C D A D : c o t e c o m m u n

△ A B D :

\frac{\sin 46}{BD} = \frac{\sin X}{AD} (1)

△ A C D

\frac{\sin 12}{CD} = \frac{\sin 62}{AD} (2)

(1) et (2) \Rightarrow \frac{BDsin X}{\sin 46} = \frac{CDsin 62}{\sin 12} (3)

△ B C D ∡ C D B = 52 - X

\frac{\sin 8}{BD} = \frac{\sin (52 - X)}{CD} (4)

(3) \times (4) \Rightarrow \frac{\sin 8 \times \sin X}{\sin 46} = \frac{\sin 62 \times \sin (52 - X)}{\sin 12}

\sin 8 \times \sin 12 \times \sin X = \sin 62 \times \sin 46 \sin (52 - X)

\sin (52 - X) = \sin 52 \sqrt{1 - \sin^{2} X} - \cos 52 \sin X

p o s i n s Z = \sin X

\frac{\sin X}{\sin (52 - X)} = \frac{\sin 62 \times \sin 46}{\sin 8 \times \sin 12}

= \frac{\sin X}{\sin 52 \cos X - \cos 52 \sin X} = \frac{\sin 62 \times \sin 46}{\sin 8 \times \sin 12}

= \frac{(\tan X)}{(\sin 52 - \cos 52 \tan X)} = \frac{\sin 62 \times \sin 46}{\sin 8 \times \sin 12}

p o s i n s t = \tan X

\frac{t}{\sin 52 - t \cos 52} = \frac{\sin 62 \times \sin 46}{\sin 8 \times \sin 12}

t (\sin 8 \sin 12 + \sin 62 \sin 46 \cos 52) = \sin 52 \sin 62 \sin 46)

t = \frac{\sin 52 \sin 62 \sin 46}{\sin 8 \sin 12 + \sin 62 \sin 46 \cos 52}

t = 1, 191753592594

X = 50 °

Commented by a.lgnaoui last updated on 01/Feb/23