A and B lie in the same horizontal plane as C, the foot of a building DC. AB=250 m, BC = 300 m and the angle of elevation of D from A is 28° AB is perpendicular to BC. (a) Calculate the height of the building (two decimal places). (b) What is the magnitude of DCB? (c) Calculate ADB (two decimal places). D 28° A C 250m 300m B

Given building DC is perpendicular to plane ABC with BC = 300 m, AB = 250 m, AB⊥BC, and ∠CAD = 28°, you want height DC, angle DCB, and angle ADB.

(a) Height

The length AC is the hypotenuse of right triangle ABC, so can be found using the Pythagorean theorem:

AC² = AB² +BC²

AC² = 250² +300² = 152,500

AC = √152500 ≈ 390.512

The height of the building can be found using the tangent relation and angle CAD.

Tan = Opposite/Adjacent

tan(CAD) = DC/AC

DC = AC·tan(CAD) = 390.512·tan(28°) ≈ 207.64

The height of the building is about 207.64 meters.

(b) Angle DCB

Building DC is perpendicular to plane ABC, so the angle DC makes with segment BC is 90°.

Angle DCB is 90°.

(c) Angle ADB

The angle between A and B as seen from point D can be computed using the tangent relation and the length of DB. Length DB is the hypotenuse of right triangle DCB, so we have ...

DB² = DC² +BC²

DB² = 207.64² +300² = 133,114.03

DB = √133114.03 ≈ 364.85

Then the desired angle is ...

tan(ADB) = AB/DB = 250/364.85

∠ADB = arctan(250/364.85)

∠ADB ≈ 34.42°