From the top of a mountain, in the same vertical plane is two consecutive kilometers stones on the level piece of ground the angles of depression to the kilometer stones are 42° 12 minutes and 23° 30 minutes respectively.
A. Calculate the hate of the mountain B. Calculate the distance to the nearest kilometre stone
Denote $\alpha = 42^{\circ}12'$, $\beta = 23^{\circ}30'$.
$h - $ height; $d - $ distance to nearest km stone.
Then $$\tan \alpha = \dfrac{h}{d}; \tag{1}$$ $$\tan \beta = \dfrac{h}{d+1}. \tag{2}$$ So : $$ \dfrac{\tan\alpha}{\tan\beta} = \dfrac{h/d}{h/(d+1)}=\dfrac{d+1}{d} =1+\dfrac{1}{d}. $$
And therefore $$ d = \dfrac{1}{\dfrac{\tan\alpha}{\tan\beta} - 1}=\dfrac{\tan\beta}{\tan\alpha-\tan\beta}.\tag{3} $$
Then $h$ can be found easily from $(1), (3)$.