Question-216369

Question Number 216369 by AROUNAMoussa last updated on 05/Feb/25

Answered by AROUNAMoussa last updated on 05/Feb/25

D e t e r m i n e : m e s I \hat{B N} p u i s m e s J \hat{M P}

Commented by a.lgnaoui last updated on 06/Feb/25

Answered by a.lgnaoui last updated on 06/Feb/25

mes (JNP) = mes (JIP) = 57

Commented by AROUNAMoussa last updated on 06/Feb/25

e t m e s (I B N) = ? m e s (J M P) = ?

Commented by a.lgnaoui last updated on 08/Feb/25

{IN}^{2} = {MI}^{2} + {MN}^{2} - 2 MN . MIcos (57 - x)

MP = PI \cos x + MI \cos (57 - x) (1)

MJ = NJ \cos x + MN \cos (57 - x) (2)

MI . MJ = MN . MP

\frac{MP}{MJ} = \frac{MI}{MN}

\frac{MP}{MJ} = \frac{PI \cos x + MI (\cos 57 \cos x + \sin 57 \sin x)}{NJ \cos x + MN (\cos 57 \cos x + \sin 57 \sin x)}

= \frac{[PI + MI (\cos 57 + \sin 57 \tan x)}{NJ + MN (\cos 57 + \sin 57 \tan x)} = \frac{MI}{MN}

MN \times PI + MN . MI (\cos 57 + \sin 57 \tan x)

= MI . NJ + MI . MN (\cos 57 + \sin 57 \tan x)

MN . PI = MI . NJ ? \frac{MI}{MN} = \frac{PI}{NJ} = \frac{MP}{MJ}

MI = \frac{MN \times PI}{NJ}

\frac{\sin x}{MI} = \sin x (\frac{\sin (57 / t - \cos 57)}{PI})

\frac{1}{MI} = \frac{\sin 57 / t - \cos 57)}{PI}

MI / PI = MN / NJ =

{MI}^{2} + {PI}^{2} - 2 MI . PI \cos 123 = {MP}^{2}

{MI}^{2} + {MI}^{2} (\sin 57 / t - \cos 57)

- 2 . {MI}^{2} . (\sin 57 / t - \cos 57) \cos 123 = {MP}^{2}

= {MI}^{2} [1 + (\sin 57 / t - \cos 57) - 2 (\sin 57 / t - \cos 57) \cos 123

{MI}^{2} [(1 + (\sin 57 / t - \cos 57) (1 - 2 \cos 123)] = {MP}^{2}

{(\frac{MP}{MI})}^{2} = [1 + (\sin 57 / t - \cos 57) (1 - \cos 123)]

\frac{\sin 123}{MP} = \frac{\sin x}{MI} \Rightarrow \frac{MP}{MI} = \frac{\sin 123}{\sin x}

donc

\frac{\sin^{2} 123}{\sin^{2} x} = (1 + \frac{\sin 57}{t} - \cos 57) (1 - \cos 123)

\frac{1}{\sin^{2} x} = 1 + \frac{1}{t^{2}}

\frac{\sin^{2} 123 (1 + t^{2})}{t^{2}} = (1 - \cos 223) (\frac{\sin 57 + t - t \cos 57}{t^{2}}) t

(1 - \cos 223) (t^{2} (1 - \cos 57) + \sin 57 t

- \sin^{2} 123 \times t^{2} - \sin^{2} 123 = 0

(1 - \cos 57) (1 - \cos 123) t^{2} + \sin 57 (1 - \cos 223) t

- \sin^{2} 123 t^{2} - \sin^{2} 123 = 0

l equation finale est :

[(1 - Cos 57) (1 - \cos 123) - \sin^{2} 123)] t^{2}

+ \sin 57 (1 - \cos 123) t - \sin^{2} 123 = 0

(1 - \cos^{2} 57 - \sin^{2} 57) t^{2} = 0

(\sin 57 + \sin 57 \cos 57) t = \sin^{2} 57

soit t = \frac{\sin 57}{(1 + \cos 57)}

x = 28, 5

\Rightarrow {\begin{cases} ∡ IBN = 57 + x = 85, 5 \\ ∡ JMP = 57 - x = 28, 5 ° \end{cases}

Commented by a.lgnaoui last updated on 08/Feb/25