A tree stands in a field Using a protractor and standing som

A tree stands in a field. Using a protractor and standing some distance from the tree one measures the angle of elevation as 48degree from the ground to the top of the tree. Moving further backward 30 ft from the tree the angle of elevation is 42degree . The height of the tree is _______ ft.

Solution

let h be height of tree

from figure

tan48^o = h/x

==> x = h/tan48^o    --------- (1)

tan42^o = h/(x + 30)

==> x + 30 = h/ tan42^o

==> x = -30 + h/ tan42^o --------- (2)

from (1) and (2)

h/tan48^o = -30 + h/ tan42^o

==> h/tan42^o - h/tan48^o = 30

==> h(1/tan42^o - 1/tan48^o) = 30

==> h (0.21021) = 30

==> h = 30/0.21021

==> h = 142.7155

Hence height of the tree = 142.7155 ft