I have not been able to solve this question of trigonometry (height and distance) in which we have to find the height of the hill when from the top of a hill, the angles of depression of two consecutive kilometer stones due east are found to be 30o and 45o.
RS Aggarwal, Class 10, Chapter 14, question no 37.
Let the distance between the nearer kilometre stone and the hill be ‘x‘ km.
So, the distance between the farther kilometre stone and the hill is ‘1+x‘ km since both are on the same side of the hill.
In triangle APB,
tan45°=h/x
⇒1=h/x
⇒h=x
In triangle AQB,
tan30°=h/(1+x)
⇒1/√3=h/(1+x)
⇒1+x=√3h
From equation 1,
1+h=√3h
⇒1=√3h−h
⇒h=1/√(3−1)
⇒h=1.365km