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Question Number 203269 by ahmetgg last updated on 13/Jan/24

	Answered by esmaeil last updated on 13/Jan/24

	$t a n 10 = \frac{A H}{B H}$ $t a n 20 = \frac{H C}{B H} \to \frac{H C}{A H} \approx 2.0642$ $t a n 50 = \frac{O H}{A H}$ $t a n x = \frac{O H}{C H} \to \frac{t a n 50}{t a n x} \approx 2.0642 \to$ $x = {t a n}^{- 1} (\frac{t a n 50}{2.0642}) \approx 30$
	Answered by mr W last updated on 14/Jan/24

	$A B = 1$ $A C = \frac{\sin 30 °}{\sin 110 °}$ $A H = \frac{\sin 10 °}{\sin 40 °}$ $A C = A H \cos 50 ° + A H \sin 50 ° \cot x$ $\frac{\sin 30 °}{\sin 110 °} = \frac{\sin 10 °}{\sin 40 °} (\cos 50 ° + \sin 50 ° \cot x)$ $\cot x = \frac{2 \cos 10 °}{\sin 50 °} - \frac{1}{\tan 50 °}$ $\Rightarrow x = 30 °$
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