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Question Number 204560 by mr W last updated on 21/Feb/24

Commented by mr W last updated on 21/Feb/24

$A C = B D$ $f i n d ∠ B = ?$
	Answered by es last updated on 21/Feb/24

	$A C = B D$ $\frac{A C}{s i n 80} = \frac{A D}{s i n 40} \to A D = \frac{A C}{2 c o s 40}$ $\frac{A D}{s i n ?} = \frac{B D}{s i n (100 - ?)} \to$ $\frac{A C}{2 c o s 40 \times s i n ?} = \frac{B D}{s i n (100 - ?)} \to$ $? = 41.53$
	Commented by A5T last updated on 21/Feb/24

	$\frac{A D}{s i n ?} = \frac{B D}{s i n (80 - ?)} \Rightarrow \frac{A C}{2 c o s 40 s i n ?} = \frac{A C}{s i n (80 - ?)}$ $\Rightarrow ? = 30 °$
	Commented by es last updated on 22/Feb/24

	$t h a n k s \underset{―}{\underset{⏟}{⋚}}$
	Answered by Ghisom last updated on 21/Feb/24

	$\frac{A C}{\sin 80 °} = \frac{A D}{\sin 40 °}$ $\frac{A C}{\sin (80 ° - β)} = \frac{A D}{\sin β}$ $- - - - - - - - -$ $this transforms to$ $\tan β = \frac{\sqrt{3}}{3}$ $β = 30 °$
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