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Question Number 181839 by amin96 last updated on 01/Dec/22

Commented by mr W last updated on 01/Dec/22

$t h e r e i s n o u n i q u e s o l u t i o n . α c a n b e$ $a n y v a l u e b e t w e e n 0 a n d 90 ° .$
Commented by Acem last updated on 02/Dec/22

$T h e i n t e n t c o u l d b e t o d i v i d e t h e s i d e b y r a t i o \frac{2}{3}$
Commented by mr W last updated on 02/Dec/22

Commented by mr W last updated on 02/Dec/22

$o n l y t w o a m o n g i n f i n i t e c a s e s$
Commented by Acem last updated on 02/Dec/22

${Y o u}^{'} r e r i g h t!$
	Answered by a.lgnaoui last updated on 02/Dec/22

	$Triangle ABD$ $\frac{\sin ∡ ABD}{3} = \frac{\sin ∡ ADB}{5} = \frac{\sin ∡ BAD}{BD}$ $\frac{\sin (α - a)}{3} = \frac{\sin (α + a)}{5}$ $5 \sin (α - a) = 3 \sin (α + a)$ $5 (\sin α \cos a - \cos α \sin a) = 3 (\sin α \cos a + \cos α \sin a)$ $2 \sin α \cos a = 8 \cos α \sin a$ $\frac{1}{4} = \tan a \times \cot α = \frac{\tan a}{\tan α} (1)$ $Triangle ABC$ ${BC}^{2} = {AB}^{2} + {AC}^{2} - 2 AB \times ACcos 2 α$ ${BC}^{2} = 50 - 50 \cos 2 α (2)$ $\tan α = \frac{AF}{BF} = \frac{2 \times AF}{BC} AF = 5 \sin α$ $\tan α = \frac{10 \sin α}{BC}$ $5^{2} = {AF}^{2} + \frac{{BC}^{2}}{4} = 5^{2} \sin^{2} α + 50 (1 - \cos 2 α)$ $\sin α = \frac{AF}{5} \Rightarrow AF = 5 \sin α$ $25 = 25 \sin^{2} α + \frac{50}{4} (1 - \cos 2 α)$ $1 = \sin^{2} α + \frac{1}{2} (1 - \cos^{2} α + \sin^{2} α)$ $1 = 2 \sin^{2} α + \frac{1}{2} \Rightarrow \sin α = \frac{1}{2} (α = \frac{π}{6})$ $\tan a = \frac{\tan α}{4} = \frac{1}{4 \sqrt{3}} = \frac{\sqrt{3}}{12}$ $A n g l e a = 8, 21^{°}$
	Commented by a.lgnaoui last updated on 02/Dec/22

	Commented by a.lgnaoui last updated on 02/Dec/22

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