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Question Number 197095 by Mingma last updated on 07/Sep/23

	Answered by mahdipoor last updated on 07/Sep/23

	$\frac{A B}{s i n 4 a} = \frac{B M}{s i n a}$ $\frac{C D}{s i n 4 a} = \frac{M C}{s i n 2 a} \Rightarrow \frac{2 . C D . c o s a}{s i n 4 a} = \frac{M C}{s i n a} \Rightarrow$ $\frac{M C}{s i n a} + \frac{B M}{s i n a} = \frac{C B}{s i n a} = \frac{2 . C D . c o s a}{s i n 4 a} + \frac{A B}{s i n 4 a}$ $\Rightarrow \frac{s i n 4 a}{s i n a} = 4 c o s 2 a . c o s a = 2 c o s a + 1 \Rightarrow$ $8 {c o s}^{3} a - 6 c o s a - 1 = 0 \Rightarrow a = 20 (h o w ? ?)$
	Commented by sniper237 last updated on 07/Sep/23

	$\overset{l i n e a r i s a t i o n}{\Rightarrow} 2 c o s (3 a) - 1 = 0$ $\overset{t r i g o e q t %}{\Rightarrow} c o s (3 a) = c o s (60)$ $T h e n 3 a = 60 a n d a = 20$
	Commented by Mingma last updated on 08/Sep/23
	Perfect ��
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