Монгол Бодлогын Сан

Заавар: $\angle HBH_1=\angle HAH_2=90^\circ-\gamma$ тул $\angle H_1HB=\angle H_2HA=\gamma$ байна. Иймд $$\begin{gather*} BH=\dfrac{BH_1}{\sin\gamma}=\dfrac{AB\cos\beta}{\sin\gamma},\\ HH_2=\dfrac{AH_2}{\tg\gamma}=\dfrac{AB\cos\alpha\cos\gamma}{\sin\gamma} \end{gather*}$$

Бодолт: $$\begin{gather*} BH=\dfrac{BH_1}{\sin\gamma}=\dfrac{AB\cos\beta}{\sin\gamma},\\ HH_2=\dfrac{AH_2}{\tg\gamma}=\dfrac{AB\cos\alpha\cos\gamma}{\sin\gamma} \end{gather*}$$ тул $$\begin{align*} BH:HH_2&=\dfrac{AB\cos\beta}{\sin\gamma}:\dfrac{AB\cos\alpha\cos\gamma}{\sin\gamma}\\ &=\dfrac{\cos\beta}{\cos\alpha\cos\gamma}=\dfrac{\cos40^\circ}{\cos60^\circ\cos80^\circ} \\ &=\dfrac{2\cos40^\circ}{\cos80^\circ}& & \color{WildStrawberry}{\leftarrow \cos60^\circ=\dfrac12}\ \end{align*}$$