The angle of elevation at the top of a building is 34 degrees. at a point 80 feet closer , the angle of elevation to the top of the same building is 45 degrees. approximate the height of the building.

According to the diagram, AB is the building.

Let distance BC= x CD= 80 ft

So distance BD= x + 80

In triangle ABD,

anlge ACB = 45 degree

So Triangle ACB is an isosceles triangle.

So AB = BC = x

Consider triangle ABD,

tan (34) =

0.6745 =

Cross multiply

0.6745 (x + 80 ) = x

0.6745 x + 53.46 = x

53.46 = x - 0.6745x

53.46 = 0.3255 x

Now divide both side by 0.3255

x = 164.24 ft

So height of the building is 164 ft approximately

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

ac ≅ ab (given)< cax ≅ < bax   (given)ax ≅ ax (common side)

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

ac ≅ ab (given)< cay ≅ < bay (since < cax ≅ < bax is given and xy is the extension of line ax)ay ≅ ay (common side)

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)

so the perpendicular line goes through the given pointyou would get a perpendicular line at some random location

step-by-step explanation:

points a and b are equidistant from the desired line, so the line you find by using them will go through point r.

using points other than a and b will give the desired line only if they are equidistant from r. if they are randomly located, the perpendicular line you create will not necessarily go through point r, hence will not be a solution to the problem.