Two trees are 120 m apart. From the point halfway between them, the angle of elevation to the top of the trees is 36 and 52. How much taller is one tree than the other.

Respuesta :

One tree is 31.044 m taller than the other one in height.

Given Information and Formula Used:

The distance between the trees, BC (from the figure) = 120 m

Elevation of angles to the top of the trees,

∠AMB = 52°

∠DMC = 36°

In a right angled triangle,

tan x = Perpendicular/Height

Here, x is the angle opposite to the Perpendicular.

Calculating the Height Difference:

Let's compute the height of the taller tree first.

In ΔAMB,

tan 52° = AB / BM

Now, since M is the point halfway between the trees,

BM = CM = BC/2

BM = CM = 60 m

⇒ 1.2799 = AB / 60

AB = 1.2799 × 60

Thus, the height of the taller tree, AB =  76.794 m

Now, we will compute the height of the smaller tree.

In ΔDCM,

tan 36° = DC / CM

⇒ 0.7625 = DC / 60

DC = 0.7625 × 60

Thus, the height of the smaller tree, DC = 45.75 m

The difference in the heights of the trees, AP = AB - DC

AP = (76.794 - 45.75)m

AP = 31.044m

Hence one tree is 31.044m taller in height than the other.

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