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.

Relax

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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